Chapter 7: Integrals

The general anti derivative of $$f(x) = 4x^3 - 6$$ is

$$F(x) = \displaystyle\int (4x^3 - 6)\,dx = 4 \cdot \dfrac{x^4}{4} - 6x + C = x^4 - 6x + C$$

We are given the condition $$F(0) = 3$$. Substituting $$x = 0$$:

$$F(0) = 0^4 - 6(0) + C = C$$

So $$C = 3$$.

Hence the required anti derivative is

$$F(x) = x^4 - 6x + 3$$

We seek a function whose derivative is $$\sin 2x$$.

Recall $$\dfrac{d}{dx}(\cos 2x) = -2\sin 2x$$.

Dividing by $$-2$$, $$\dfrac{d}{dx}\left(-\dfrac{1}{2}\cos 2x\right) = \sin 2x$$.

So $$-\dfrac{1}{2}\cos 2x$$ is one anti derivative. Including the arbitrary constant of integration, the general anti derivative of $$\sin 2x$$ is $$-\dfrac{1}{2}\cos 2x + C$$.

We seek a function whose derivative is $$\cos 3x$$.

Recall $$\dfrac{d}{dx}(\sin 3x) = 3\cos 3x$$.

Dividing by $$3$$, $$\dfrac{d}{dx}\left(\dfrac{1}{3}\sin 3x\right) = \cos 3x$$.

So $$\dfrac{1}{3}\sin 3x$$ is one anti derivative. Including the arbitrary constant of integration, the general anti derivative of $$\cos 3x$$ is $$\dfrac{1}{3}\sin 3x + C$$.

We seek a function whose derivative is $$e^{2x}$$.

Recall $$\dfrac{d}{dx}(e^{2x}) = 2e^{2x}$$.

Dividing by $$2$$, $$\dfrac{d}{dx}\left(\dfrac{1}{2}e^{2x}\right) = e^{2x}$$.

So $$\dfrac{1}{2}e^{2x}$$ is one anti derivative. Including the arbitrary constant of integration, the general anti derivative of $$e^{2x}$$ is $$\dfrac{1}{2}e^{2x} + C$$.

We seek a function whose derivative is $$(ax + b)^2$$.

Recall, by the chain rule, $$\dfrac{d}{dx}(ax + b)^3 = 3(ax + b)^2 \cdot a = 3a(ax + b)^2$$.

Dividing by $$3a$$ (with $$a \neq 0$$), $$\dfrac{d}{dx}\left(\dfrac{(ax + b)^3}{3a}\right) = (ax + b)^2$$.

So $$\dfrac{(ax + b)^3}{3a}$$ is one anti derivative. Including the arbitrary constant of integration, the general anti derivative of $$(ax + b)^2$$ is $$\dfrac{(ax + b)^3}{3a} + C$$.

We find an anti derivative of each term separately.

For $$\sin 2x$$: since $$\dfrac{d}{dx}(\cos 2x) = -2\sin 2x$$, an anti derivative is $$-\dfrac{1}{2}\cos 2x$$.

For $$4e^{3x}$$: since $$\dfrac{d}{dx}(e^{3x}) = 3e^{3x}$$, an anti derivative of $$4e^{3x}$$ is $$\dfrac{4}{3}e^{3x}$$.

Combining the two and including the arbitrary constant of integration, the general anti derivative of $$\sin 2x - 4e^{3x}$$ is

$$-\dfrac{1}{2}\cos 2x - \dfrac{4}{3}e^{3x} + C$$

Integrate term by term:

$$\displaystyle\int (4 e^{3x} + 1)\,dx = 4\int e^{3x}\,dx + \int 1\,dx$$

Using $$\displaystyle\int e^{3x}\,dx = \dfrac{e^{3x}}{3}$$:

$$= 4 \cdot \dfrac{e^{3x}}{3} + x + C = \dfrac{4}{3}e^{3x} + x + C$$

First multiply out the integrand:

$$x^2\left(1 - \dfrac{1}{x^2}\right) = x^2 - 1$$

Therefore

$$\displaystyle\int x^2\left(1 - \dfrac{1}{x^2}\right) dx = \int (x^2 - 1)\,dx = \dfrac{x^3}{3} - x + C$$

Here $$a, b, c$$ are constants. Integrate term by term:

$$\displaystyle\int (ax^2 + bx + c)\,dx = a\int x^2\,dx + b\int x\,dx + c\int 1\,dx$$

$$= a \cdot \dfrac{x^3}{3} + b \cdot \dfrac{x^2}{2} + cx + C$$

$$= \dfrac{ax^3}{3} + \dfrac{bx^2}{2} + cx + C$$

Integrate term by term:

$$\displaystyle\int (2x^2 + e^x)\,dx = 2\int x^2\,dx + \int e^x\,dx$$

$$= 2 \cdot \dfrac{x^3}{3} + e^x + C = \dfrac{2x^3}{3} + e^x + C$$

Expand the square using $$(a - b)^2 = a^2 - 2ab + b^2$$:

$$\left(\sqrt{x} - \dfrac{1}{\sqrt{x}}\right)^2 = x - 2\sqrt{x}\cdot\dfrac{1}{\sqrt{x}} + \dfrac{1}{x} = x - 2 + \dfrac{1}{x}$$

Therefore

$$\displaystyle\int \left(\sqrt{x} - \dfrac{1}{\sqrt{x}}\right)^2 dx = \int \left(x - 2 + \dfrac{1}{x}\right) dx$$

$$= \dfrac{x^2}{2} - 2x + \log |x| + C$$

Divide each term of the numerator by $$x^2$$:

$$\dfrac{x^3 + 5x^2 - 4}{x^2} = x + 5 - 4x^{-2}$$

Therefore

$$\displaystyle\int \dfrac{x^3 + 5x^2 - 4}{x^2}\,dx = \int \left(x + 5 - 4x^{-2}\right) dx$$

$$= \dfrac{x^2}{2} + 5x - 4 \cdot \dfrac{x^{-1}}{-1} + C = \dfrac{x^2}{2} + 5x + \dfrac{4}{x} + C$$

Divide each term of the numerator by $$\sqrt{x} = x^{1/2}$$:

$$\dfrac{x^3 + 3x + 4}{\sqrt{x}} = x^{5/2} + 3x^{1/2} + 4x^{-1/2}$$

Therefore

$$\displaystyle\int \dfrac{x^3 + 3x + 4}{\sqrt{x}}\,dx = \int \left(x^{5/2} + 3x^{1/2} + 4x^{-1/2}\right) dx$$

$$= \dfrac{x^{7/2}}{7/2} + 3 \cdot \dfrac{x^{3/2}}{3/2} + 4 \cdot \dfrac{x^{1/2}}{1/2} + C$$

$$= \dfrac{2}{7}x^{7/2} + 2x^{3/2} + 8\sqrt{x} + C$$

Factorise the numerator by grouping:

$$x^3 - x^2 + x - 1 = x^2(x - 1) + 1(x - 1) = (x - 1)(x^2 + 1)$$

So, for $$x \neq 1$$,

$$\dfrac{x^3 - x^2 + x - 1}{x - 1} = x^2 + 1$$

Therefore

$$\displaystyle\int \dfrac{x^3 - x^2 + x - 1}{x - 1}\,dx = \int (x^2 + 1)\,dx = \dfrac{x^3}{3} + x + C$$

Multiply out the integrand:

$$(1 - x)\sqrt{x} = \sqrt{x} - x\sqrt{x} = x^{1/2} - x^{3/2}$$

Therefore

$$\displaystyle\int (1 - x)\sqrt{x}\,dx = \int \left(x^{1/2} - x^{3/2}\right) dx$$

$$= \dfrac{x^{3/2}}{3/2} - \dfrac{x^{5/2}}{5/2} + C = \dfrac{2}{3}x^{3/2} - \dfrac{2}{5}x^{5/2} + C$$

Multiply out the integrand:

$$\sqrt{x}\,(3x^2 + 2x + 3) = 3x^{5/2} + 2x^{3/2} + 3x^{1/2}$$

Therefore

$$\displaystyle\int \sqrt{x}\,(3x^2 + 2x + 3)\,dx = \int \left(3x^{5/2} + 2x^{3/2} + 3x^{1/2}\right) dx$$

$$= 3 \cdot \dfrac{x^{7/2}}{7/2} + 2 \cdot \dfrac{x^{5/2}}{5/2} + 3 \cdot \dfrac{x^{3/2}}{3/2} + C$$

$$= \dfrac{6}{7}x^{7/2} + \dfrac{4}{5}x^{5/2} + 2x^{3/2} + C$$

Integrate term by term:

$$\displaystyle\int (2x - 3\cos x + e^x)\,dx = 2\int x\,dx - 3\int \cos x\,dx + \int e^x\,dx$$

$$= 2 \cdot \dfrac{x^2}{2} - 3\sin x + e^x + C = x^2 - 3\sin x + e^x + C$$

Integrate term by term, writing $$\sqrt{x} = x^{1/2}$$:

$$\displaystyle\int (2x^2 - 3\sin x + 5\sqrt{x})\,dx = 2\int x^2\,dx - 3\int \sin x\,dx + 5\int x^{1/2}\,dx$$

$$= 2 \cdot \dfrac{x^3}{3} - 3(-\cos x) + 5 \cdot \dfrac{x^{3/2}}{3/2} + C$$

$$= \dfrac{2x^3}{3} + 3\cos x + \dfrac{10}{3}x^{3/2} + C$$

Expand the integrand:

$$\sec x\,(\sec x + \tan x) = \sec^2 x + \sec x \tan x$$

Now integrate, using $$\displaystyle\int \sec^2 x\,dx = \tan x$$ and $$\displaystyle\int \sec x \tan x\,dx = \sec x$$:

$$\displaystyle\int \sec x\,(\sec x + \tan x)\,dx = \tan x + \sec x + C$$

Simplify the integrand:

$$\dfrac{\sec^2 x}{\operatorname{cosec}^2 x} = \dfrac{1/\cos^2 x}{1/\sin^2 x} = \dfrac{\sin^2 x}{\cos^2 x} = \tan^2 x$$

Using the identity $$\tan^2 x = \sec^2 x - 1$$:

$$\displaystyle\int \dfrac{\sec^2 x}{\operatorname{cosec}^2 x}\,dx = \int (\sec^2 x - 1)\,dx = \tan x - x + C$$

Split the integrand into two fractions:

$$\dfrac{2 - 3\sin x}{\cos^2 x} = \dfrac{2}{\cos^2 x} - \dfrac{3\sin x}{\cos^2 x} = 2\sec^2 x - 3\sec x \tan x$$

Now integrate:

$$\displaystyle\int \dfrac{2 - 3\sin x}{\cos^2 x}\,dx = 2\int \sec^2 x\,dx - 3\int \sec x \tan x\,dx$$

$$= 2\tan x - 3\sec x + C$$

Write the function in index form: $$\sqrt{x} + \dfrac{1}{\sqrt{x}} = x^{1/2} + x^{-1/2}$$.

The anti derivative is

$$\displaystyle\int \left(x^{1/2} + x^{-1/2}\right) dx = \dfrac{x^{3/2}}{3/2} + \dfrac{x^{1/2}}{1/2} + C$$

$$= \dfrac{2}{3}x^{3/2} + 2x^{1/2} + C$$

This matches option (C).

We are given $$\dfrac{d}{dx}f(x) = 4x^3 - 3x^{-4}$$, so $$f(x)$$ is the anti derivative:

$$f(x) = \displaystyle\int \left(4x^3 - 3x^{-4}\right) dx = 4 \cdot \dfrac{x^4}{4} - 3 \cdot \dfrac{x^{-3}}{-3} + C$$

$$f(x) = x^4 + x^{-3} + C = x^4 + \dfrac{1}{x^3} + C$$

Apply the condition $$f(2) = 0$$:

$$f(2) = 2^4 + \dfrac{1}{2^3} + C = 16 + \dfrac{1}{8} + C = 0$$

$$\Rightarrow C = -16 - \dfrac{1}{8} = -\dfrac{129}{8}$$

Hence $$f(x) = x^4 + \dfrac{1}{x^3} - \dfrac{129}{8}$$, which is option (A).

Put $$t = 1 + x^2$$, so $$dt = 2x\,dx$$.

$$\displaystyle\int \dfrac{2x}{1 + x^2}\,dx = \int \dfrac{dt}{t} = \log |t| + C$$

Since $$1 + x^2 > 0$$,

$$= \log(1 + x^2) + C$$

Put $$t = \log x$$, so $$dt = \dfrac{1}{x}\,dx$$.

$$\displaystyle\int \dfrac{(\log x)^2}{x}\,dx = \int t^2\,dt = \dfrac{t^3}{3} + C$$

$$= \dfrac{(\log x)^3}{3} + C$$

Factor the denominator: $$x + x\log x = x(1 + \log x)$$.

$$\displaystyle\int \dfrac{1}{x(1 + \log x)}\,dx$$

Put $$t = 1 + \log x$$, so $$dt = \dfrac{1}{x}\,dx$$:

$$= \int \dfrac{dt}{t} = \log |t| + C = \log |1 + \log x| + C$$

Put $$t = \cos x$$, so $$dt = -\sin x\,dx$$, i.e. $$\sin x\,dx = -dt$$.

$$\displaystyle\int \sin x\,\sin(\cos x)\,dx = \int \sin t\,(-dt) = -\int \sin t\,dt$$

$$= -(-\cos t) + C = \cos t + C = \cos(\cos x) + C$$

Use the identity $$2\sin\theta\cos\theta = \sin 2\theta$$:

$$\sin(ax + b)\cos(ax + b) = \dfrac{1}{2}\sin\big(2(ax + b)\big)$$

So

$$I = \dfrac{1}{2}\int \sin(2ax + 2b)\,dx$$

Put $$t = 2ax + 2b$$, $$dt = 2a\,dx$$:

$$I = \dfrac{1}{2} \cdot \dfrac{1}{2a}\int \sin t\,dt = \dfrac{1}{4a}(-\cos t) + C$$

$$= -\dfrac{1}{4a}\cos(2ax + 2b) + C$$

Put $$t = ax + b$$, so $$dt = a\,dx$$, i.e. $$dx = \dfrac{dt}{a}$$.

$$\displaystyle\int \sqrt{ax + b}\,dx = \int t^{1/2} \cdot \dfrac{dt}{a} = \dfrac{1}{a} \cdot \dfrac{t^{3/2}}{3/2} + C$$

$$= \dfrac{2}{3a}t^{3/2} + C = \dfrac{2}{3a}(ax + b)^{3/2} + C$$

Put $$t = x + 2$$, so $$x = t - 2$$ and $$dx = dt$$.

$$\displaystyle\int x\sqrt{x + 2}\,dx = \int (t - 2)\sqrt{t}\,dt = \int \left(t^{3/2} - 2t^{1/2}\right) dt$$

$$= \dfrac{t^{5/2}}{5/2} - 2 \cdot \dfrac{t^{3/2}}{3/2} + C = \dfrac{2}{5}t^{5/2} - \dfrac{4}{3}t^{3/2} + C$$

$$= \dfrac{2}{5}(x + 2)^{5/2} - \dfrac{4}{3}(x + 2)^{3/2} + C$$

Put $$t = 1 + 2x^2$$, so $$dt = 4x\,dx$$, i.e. $$x\,dx = \dfrac{dt}{4}$$.

$$\displaystyle\int x\sqrt{1 + 2x^2}\,dx = \int \sqrt{t} \cdot \dfrac{dt}{4} = \dfrac{1}{4} \cdot \dfrac{t^{3/2}}{3/2} + C$$

$$= \dfrac{1}{6}t^{3/2} + C = \dfrac{1}{6}(1 + 2x^2)^{3/2} + C$$

Note that $$4x + 2 = 2(2x + 1)$$ and $$\dfrac{d}{dx}(x^2 + x + 1) = 2x + 1$$.

Put $$t = x^2 + x + 1$$, so $$dt = (2x + 1)\,dx$$.

$$\displaystyle\int (4x + 2)\sqrt{x^2 + x + 1}\,dx = \int 2\sqrt{t}\,dt = 2 \cdot \dfrac{t^{3/2}}{3/2} + C$$

$$= \dfrac{4}{3}t^{3/2} + C = \dfrac{4}{3}(x^2 + x + 1)^{3/2} + C$$

Put $$t = \sqrt{x}$$, so $$x = t^2$$ and $$dx = 2t\,dt$$.

$$\displaystyle\int \dfrac{1}{x - \sqrt{x}}\,dx = \int \dfrac{2t\,dt}{t^2 - t} = \int \dfrac{2t\,dt}{t(t - 1)} = \int \dfrac{2\,dt}{t - 1}$$

$$= 2\log |t - 1| + C = 2\log |\sqrt{x} - 1| + C$$

Put $$t = x + 4$$, so $$x = t - 4$$ and $$dx = dt$$.

$$\displaystyle\int \dfrac{x}{\sqrt{x + 4}}\,dx = \int \dfrac{t - 4}{\sqrt{t}}\,dt = \int \left(t^{1/2} - 4t^{-1/2}\right) dt$$

$$= \dfrac{t^{3/2}}{3/2} - 4 \cdot \dfrac{t^{1/2}}{1/2} + C = \dfrac{2}{3}t^{3/2} - 8\sqrt{t} + C$$

$$= \dfrac{2}{3}(x + 4)^{3/2} - 8\sqrt{x + 4} + C$$

Put $$t = x^3 - 1$$, so $$x^3 = t + 1$$ and $$dt = 3x^2\,dx$$, i.e. $$x^2\,dx = \dfrac{dt}{3}$$.

Write $$x^5\,dx = x^3 \cdot x^2\,dx = (t + 1)\dfrac{dt}{3}$$.

$$\displaystyle\int (x^3 - 1)^{1/3} x^5\,dx = \dfrac{1}{3}\int t^{1/3}(t + 1)\,dt = \dfrac{1}{3}\int \left(t^{4/3} + t^{1/3}\right) dt$$

$$= \dfrac{1}{3}\left[\dfrac{t^{7/3}}{7/3} + \dfrac{t^{4/3}}{4/3}\right] + C = \dfrac{1}{7}t^{7/3} + \dfrac{1}{4}t^{4/3} + C$$

$$= \dfrac{1}{7}(x^3 - 1)^{7/3} + \dfrac{1}{4}(x^3 - 1)^{4/3} + C$$

Put $$t = 2 + 3x^3$$, so $$dt = 9x^2\,dx$$, i.e. $$x^2\,dx = \dfrac{dt}{9}$$.

$$\displaystyle\int \dfrac{x^2}{(2 + 3x^3)^3}\,dx = \dfrac{1}{9}\int \dfrac{dt}{t^3} = \dfrac{1}{9}\int t^{-3}\,dt$$

$$= \dfrac{1}{9} \cdot \dfrac{t^{-2}}{-2} + C = -\dfrac{1}{18 t^2} + C = -\dfrac{1}{18(2 + 3x^3)^2} + C$$

Put $$t = \log x$$, so $$dt = \dfrac{1}{x}\,dx$$.

$$\displaystyle\int \dfrac{1}{x (\log x)^m}\,dx = \int t^{-m}\,dt = \dfrac{t^{-m + 1}}{-m + 1} + C$$

$$= \dfrac{(\log x)^{1 - m}}{1 - m} + C$$

Put $$t = 9 - 4x^2$$, so $$dt = -8x\,dx$$, i.e. $$x\,dx = -\dfrac{dt}{8}$$.

$$\displaystyle\int \dfrac{x}{9 - 4x^2}\,dx = \int \dfrac{1}{t}\left(-\dfrac{dt}{8}\right) = -\dfrac{1}{8}\int \dfrac{dt}{t}$$

$$= -\dfrac{1}{8}\log |t| + C = -\dfrac{1}{8}\log |9 - 4x^2| + C$$

Put $$t = 2x + 3$$, so $$dt = 2\,dx$$, i.e. $$dx = \dfrac{dt}{2}$$.

$$\displaystyle\int e^{2x + 3}\,dx = \int e^t \cdot \dfrac{dt}{2} = \dfrac{1}{2}e^t + C$$

$$= \dfrac{1}{2}e^{2x + 3} + C$$

Write $$\dfrac{x}{e^{x^2}} = x\,e^{-x^2}$$.

Put $$t = -x^2$$, so $$dt = -2x\,dx$$, i.e. $$x\,dx = -\dfrac{dt}{2}$$.

$$\displaystyle\int x\,e^{-x^2}\,dx = \int e^t\left(-\dfrac{dt}{2}\right) = -\dfrac{1}{2}e^t + C$$

$$= -\dfrac{1}{2}e^{-x^2} + C$$

Put $$t = \tan^{-1} x$$, so $$dt = \dfrac{dx}{1 + x^2}$$.

$$\displaystyle\int \dfrac{e^{\tan^{-1} x}}{1 + x^2}\,dx = \int e^t\,dt = e^t + C$$

$$= e^{\tan^{-1} x} + C$$

Divide numerator and denominator by $$e^x$$:

$$\dfrac{e^{2x} - 1}{e^{2x} + 1} = \dfrac{e^x - e^{-x}}{e^x + e^{-x}}$$

Put $$t = e^x + e^{-x}$$, so $$dt = (e^x - e^{-x})\,dx$$.

$$\displaystyle\int \dfrac{e^x - e^{-x}}{e^x + e^{-x}}\,dx = \int \dfrac{dt}{t} = \log |t| + C$$

Since $$e^x + e^{-x} > 0$$,

$$= \log(e^x + e^{-x}) + C$$

Put $$t = e^{2x} + e^{-2x}$$. Then

$$dt = (2e^{2x} - 2e^{-2x})\,dx = 2(e^{2x} - e^{-2x})\,dx$$

So $$(e^{2x} - e^{-2x})\,dx = \dfrac{dt}{2}$$, and

$$\displaystyle\int \dfrac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}}\,dx = \int \dfrac{1}{t} \cdot \dfrac{dt}{2} = \dfrac{1}{2}\log |t| + C$$

$$= \dfrac{1}{2}\log(e^{2x} + e^{-2x}) + C$$

Use the identity $$\tan^2\theta = \sec^2\theta - 1$$:

$$\tan^2(2x - 3) = \sec^2(2x - 3) - 1$$

Therefore

$$\displaystyle\int \tan^2(2x - 3)\,dx = \int \sec^2(2x - 3)\,dx - \int 1\,dx$$

$$= \dfrac{1}{2}\tan(2x - 3) - x + C$$

Put $$t = 7 - 4x$$, so $$dt = -4\,dx$$, i.e. $$dx = -\dfrac{dt}{4}$$.

$$\displaystyle\int \sec^2(7 - 4x)\,dx = \int \sec^2 t\left(-\dfrac{dt}{4}\right) = -\dfrac{1}{4}\tan t + C$$

$$= -\dfrac{1}{4}\tan(7 - 4x) + C$$

Put $$t = \sin^{-1} x$$, so $$dt = \dfrac{dx}{\sqrt{1 - x^2}}$$.

$$\displaystyle\int \dfrac{\sin^{-1} x}{\sqrt{1 - x^2}}\,dx = \int t\,dt = \dfrac{t^2}{2} + C$$

$$= \dfrac{(\sin^{-1} x)^2}{2} + C$$

Let the denominator be $$t = 6\cos x + 4\sin x$$. Then

$$dt = (-6\sin x + 4\cos x)\,dx = (4\cos x - 6\sin x)\,dx = 2(2\cos x - 3\sin x)\,dx$$

So the numerator $$(2\cos x - 3\sin x)\,dx = \dfrac{dt}{2}$$, giving

$$\displaystyle\int \dfrac{2\cos x - 3\sin x}{6\cos x + 4\sin x}\,dx = \int \dfrac{1}{t} \cdot \dfrac{dt}{2} = \dfrac{1}{2}\log |t| + C$$

$$= \dfrac{1}{2}\log |6\cos x + 4\sin x| + C$$

Since $$\dfrac{1}{\cos^2 x} = \sec^2 x$$, the integrand is $$\dfrac{\sec^2 x}{(1 - \tan x)^2}$$.

Put $$t = 1 - \tan x$$, so $$dt = -\sec^2 x\,dx$$.

$$\displaystyle\int \dfrac{\sec^2 x}{(1 - \tan x)^2}\,dx = \int \dfrac{-dt}{t^2} = -\int t^{-2}\,dt$$

$$= -\dfrac{t^{-1}}{-1} + C = \dfrac{1}{t} + C = \dfrac{1}{1 - \tan x} + C$$

Put $$t = \sqrt{x}$$, so $$dt = \dfrac{1}{2\sqrt{x}}\,dx$$, i.e. $$\dfrac{dx}{\sqrt{x}} = 2\,dt$$.

$$\displaystyle\int \dfrac{\cos \sqrt{x}}{\sqrt{x}}\,dx = \int \cos t \cdot 2\,dt = 2\sin t + C$$

$$= 2\sin \sqrt{x} + C$$

Put $$t = \sin 2x$$, so $$dt = 2\cos 2x\,dx$$, i.e. $$\cos 2x\,dx = \dfrac{dt}{2}$$.

$$\displaystyle\int \sqrt{\sin 2x}\,\cos 2x\,dx = \int \sqrt{t} \cdot \dfrac{dt}{2} = \dfrac{1}{2} \cdot \dfrac{t^{3/2}}{3/2} + C$$

$$= \dfrac{1}{3}t^{3/2} + C = \dfrac{1}{3}(\sin 2x)^{3/2} + C$$

Put $$t = 1 + \sin x$$, so $$dt = \cos x\,dx$$.

$$\displaystyle\int \dfrac{\cos x}{\sqrt{1 + \sin x}}\,dx = \int \dfrac{dt}{\sqrt{t}} = \int t^{-1/2}\,dt$$

$$= 2\sqrt{t} + C = 2\sqrt{1 + \sin x} + C$$

Put $$t = \log \sin x$$. Then

$$dt = \dfrac{1}{\sin x} \cdot \cos x\,dx = \cot x\,dx$$

Therefore

$$\displaystyle\int \cot x\,\log \sin x\,dx = \int t\,dt = \dfrac{t^2}{2} + C$$

$$= \dfrac{(\log \sin x)^2}{2} + C$$

Put $$t = 1 + \cos x$$, so $$dt = -\sin x\,dx$$, i.e. $$\sin x\,dx = -dt$$.

$$\displaystyle\int \dfrac{\sin x}{1 + \cos x}\,dx = \int \dfrac{-dt}{t} = -\log |t| + C$$

$$= -\log |1 + \cos x| + C$$

Put $$t = 1 + \cos x$$, so $$dt = -\sin x\,dx$$, i.e. $$\sin x\,dx = -dt$$.

$$\displaystyle\int \dfrac{\sin x}{(1 + \cos x)^2}\,dx = \int \dfrac{-dt}{t^2} = -\int t^{-2}\,dt$$

$$= -\dfrac{t^{-1}}{-1} + C = \dfrac{1}{t} + C = \dfrac{1}{1 + \cos x} + C$$

Write $$\cot x = \dfrac{\cos x}{\sin x}$$:

$$\dfrac{1}{1 + \cot x} = \dfrac{\sin x}{\sin x + \cos x}$$

Express the numerator using the denominator and its derivative:

$$\sin x = \tfrac{1}{2}\big[(\sin x + \cos x) + (\sin x - \cos x)\big]$$

So

$$I = \dfrac{1}{2}\int dx + \dfrac{1}{2}\int \dfrac{\sin x - \cos x}{\sin x + \cos x}\,dx$$

In the second integral, put $$u = \sin x + \cos x$$, $$du = (\cos x - \sin x)\,dx = -(\sin x - \cos x)\,dx$$, which gives $$-\log |u|$$.

$$I = \dfrac{x}{2} - \dfrac{1}{2}\log |\sin x + \cos x| + C$$

Write $$\tan x = \dfrac{\sin x}{\cos x}$$:

$$\dfrac{1}{1 - \tan x} = \dfrac{\cos x}{\cos x - \sin x}$$

Express the numerator using the denominator and its derivative:

$$\cos x = \tfrac{1}{2}\big[(\cos x - \sin x) + (\cos x + \sin x)\big]$$

So

$$I = \dfrac{1}{2}\int dx + \dfrac{1}{2}\int \dfrac{\cos x + \sin x}{\cos x - \sin x}\,dx$$

In the second integral, put $$u = \cos x - \sin x$$, $$du = (-\sin x - \cos x)\,dx = -(\sin x + \cos x)\,dx$$, which gives $$-\log |u|$$.

$$I = \dfrac{x}{2} - \dfrac{1}{2}\log |\cos x - \sin x| + C$$

Divide numerator and denominator by $$\cos^2 x$$. Since $$\dfrac{\sin x \cos x}{\cos^2 x} = \tan x$$:

$$\dfrac{\sqrt{\tan x}}{\sin x \cos x} = \dfrac{\sqrt{\tan x}\,\sec^2 x}{\tan x} = \dfrac{\sec^2 x}{\sqrt{\tan x}}$$

Put $$t = \tan x$$, so $$dt = \sec^2 x\,dx$$:

$$\displaystyle\int \dfrac{\sec^2 x}{\sqrt{\tan x}}\,dx = \int \dfrac{dt}{\sqrt{t}} = 2\sqrt{t} + C$$

$$= 2\sqrt{\tan x} + C$$

Put $$t = 1 + \log x$$, so $$dt = \dfrac{1}{x}\,dx$$.

$$\displaystyle\int \dfrac{(1 + \log x)^2}{x}\,dx = \int t^2\,dt = \dfrac{t^3}{3} + C$$

$$= \dfrac{(1 + \log x)^3}{3} + C$$

Write $$\dfrac{x + 1}{x} = 1 + \dfrac{1}{x}$$, so the integrand is

$$\left(1 + \dfrac{1}{x}\right)(x + \log x)^2$$

Put $$t = x + \log x$$. Then $$dt = \left(1 + \dfrac{1}{x}\right) dx$$.

$$\displaystyle\int \left(1 + \dfrac{1}{x}\right)(x + \log x)^2\,dx = \int t^2\,dt = \dfrac{t^3}{3} + C$$

$$= \dfrac{(x + \log x)^3}{3} + C$$

Put $$t = \tan^{-1}(x^4)$$. Then

$$dt = \dfrac{4x^3}{1 + (x^4)^2}\,dx = \dfrac{4x^3}{1 + x^8}\,dx \quad\Rightarrow\quad \dfrac{x^3\,dx}{1 + x^8} = \dfrac{dt}{4}$$

Therefore

$$\displaystyle\int \dfrac{x^3 \sin(\tan^{-1} x^4)}{1 + x^8}\,dx = \int \sin t \cdot \dfrac{dt}{4} = -\dfrac{1}{4}\cos t + C$$

$$= -\dfrac{1}{4}\cos(\tan^{-1} x^4) + C$$

Let $$t = x^{10} + 10^x$$ be the denominator. Differentiating:

$$\dfrac{dt}{dx} = 10x^9 + 10^x \log_e 10$$

This is exactly the numerator. So $$dt = (10x^9 + 10^x \log_e 10)\,dx$$, and

$$\displaystyle\int \dfrac{10 x^9 + 10^x \log_e 10}{x^{10} + 10^x}\,dx = \int \dfrac{dt}{t} = \log |t| + C$$

$$= \log(x^{10} + 10^x) + C$$

This matches option (D).

Use $$\sin^2 x + \cos^2 x = 1$$ in the numerator:

$$\dfrac{1}{\sin^2 x\,\cos^2 x} = \dfrac{\sin^2 x + \cos^2 x}{\sin^2 x\,\cos^2 x} = \dfrac{1}{\cos^2 x} + \dfrac{1}{\sin^2 x} = \sec^2 x + \operatorname{cosec}^2 x$$

Therefore

$$\displaystyle\int \dfrac{dx}{\sin^2 x\,\cos^2 x} = \int (\sec^2 x + \operatorname{cosec}^2 x)\,dx = \tan x - \cot x + C$$

This matches option (B).

Use $$\sin^2\theta = \dfrac{1 - \cos 2\theta}{2}$$ with $$\theta = 2x + 5$$:

$$\sin^2(2x + 5) = \dfrac{1 - \cos(4x + 10)}{2}$$

Therefore

$$\displaystyle\int \sin^2(2x + 5)\,dx = \dfrac{1}{2}\int \big[1 - \cos(4x + 10)\big]\,dx$$

$$= \dfrac{1}{2}\left(x - \dfrac{\sin(4x + 10)}{4}\right) + C = \dfrac{x}{2} - \dfrac{\sin(4x + 10)}{8} + C$$

Use $$\sin A \cos B = \dfrac{1}{2}\big[\sin(A + B) + \sin(A - B)\big]$$:

$$\sin 3x \cos 4x = \dfrac{1}{2}\big[\sin 7x + \sin(-x)\big] = \dfrac{1}{2}\big[\sin 7x - \sin x\big]$$

Therefore

$$\displaystyle\int \sin 3x \cos 4x\,dx = \dfrac{1}{2}\int (\sin 7x - \sin x)\,dx$$

$$= \dfrac{1}{2}\left(-\dfrac{\cos 7x}{7} + \cos x\right) + C = -\dfrac{\cos 7x}{14} + \dfrac{\cos x}{2} + C$$

First combine $$\cos 2x \cos 4x$$ using $$\cos A \cos B = \dfrac{1}{2}\big[\cos(A + B) + \cos(A - B)\big]$$:

$$\cos 2x \cos 4x = \dfrac{1}{2}(\cos 6x + \cos 2x)$$

Multiply by $$\cos 6x$$:

$$\cos 2x \cos 4x \cos 6x = \dfrac{1}{2}\big(\cos^2 6x + \cos 2x \cos 6x\big)$$

Now $$\cos^2 6x = \dfrac{1 + \cos 12x}{2}$$ and $$\cos 2x \cos 6x = \dfrac{1}{2}(\cos 8x + \cos 4x)$$, so

$$\cos 2x \cos 4x \cos 6x = \dfrac{1}{4}\big(1 + \cos 12x + \cos 8x + \cos 4x\big)$$

Integrating term by term:

$$\displaystyle\int \cos 2x \cos 4x \cos 6x\,dx = \dfrac{1}{4}\left(x + \dfrac{\sin 12x}{12} + \dfrac{\sin 8x}{8} + \dfrac{\sin 4x}{4}\right) + C$$

Use $$\sin^3\theta = \dfrac{3\sin\theta - \sin 3\theta}{4}$$ with $$\theta = 2x + 1$$:

$$\sin^3(2x + 1) = \dfrac{3\sin(2x + 1) - \sin(6x + 3)}{4}$$

Therefore

$$\displaystyle\int \sin^3(2x + 1)\,dx = \dfrac{1}{4}\int \big[3\sin(2x + 1) - \sin(6x + 3)\big]\,dx$$

$$= \dfrac{1}{4}\left[-\dfrac{3\cos(2x + 1)}{2} + \dfrac{\cos(6x + 3)}{6}\right] + C$$

$$= -\dfrac{3}{8}\cos(2x + 1) + \dfrac{1}{24}\cos(6x + 3) + C$$

Write $$\cos^3 x = \cos^2 x \cdot \cos x = (1 - \sin^2 x)\cos x$$, so

$$\sin^3 x \cos^3 x = \sin^3 x\,(1 - \sin^2 x)\cos x$$

Put $$t = \sin x$$, so $$dt = \cos x\,dx$$:

$$\displaystyle\int \sin^3 x \cos^3 x\,dx = \int t^3 (1 - t^2)\,dt = \int (t^3 - t^5)\,dt$$

$$= \dfrac{t^4}{4} - \dfrac{t^6}{6} + C = \dfrac{\sin^4 x}{4} - \dfrac{\sin^6 x}{6} + C$$

First combine $$\sin x \sin 3x$$ using $$\sin A \sin B = \dfrac{1}{2}\big[\cos(A - B) - \cos(A + B)\big]$$:

$$\sin x \sin 3x = \dfrac{1}{2}(\cos 2x - \cos 4x)$$

Multiply by $$\sin 2x$$:

$$\sin x \sin 2x \sin 3x = \dfrac{1}{2}\big(\sin 2x \cos 2x - \sin 2x \cos 4x\big)$$

Using $$\sin 2x \cos 2x = \dfrac{1}{2}\sin 4x$$ and $$\sin 2x \cos 4x = \dfrac{1}{2}(\sin 6x - \sin 2x)$$:

$$\sin x \sin 2x \sin 3x = \dfrac{1}{4}\big(\sin 4x + \sin 2x - \sin 6x\big)$$

Integrating:

$$\displaystyle\int \sin x \sin 2x \sin 3x\,dx = \dfrac{1}{4}\left(-\dfrac{\cos 4x}{4} - \dfrac{\cos 2x}{2} + \dfrac{\cos 6x}{6}\right) + C$$

Use $$\sin A \sin B = \dfrac{1}{2}\big[\cos(A - B) - \cos(A + B)\big]$$:

$$\sin 4x \sin 8x = \dfrac{1}{2}(\cos 4x - \cos 12x)$$

Therefore

$$\displaystyle\int \sin 4x \sin 8x\,dx = \dfrac{1}{2}\int (\cos 4x - \cos 12x)\,dx$$

$$= \dfrac{1}{2}\left(\dfrac{\sin 4x}{4} - \dfrac{\sin 12x}{12}\right) + C = \dfrac{\sin 4x}{8} - \dfrac{\sin 12x}{24} + C$$

Use the half-angle identities $$1 - \cos x = 2\sin^2\dfrac{x}{2}$$ and $$1 + \cos x = 2\cos^2\dfrac{x}{2}$$:

$$\dfrac{1 - \cos x}{1 + \cos x} = \dfrac{2\sin^2(x/2)}{2\cos^2(x/2)} = \tan^2\dfrac{x}{2} = \sec^2\dfrac{x}{2} - 1$$

Therefore

$$\displaystyle\int \dfrac{1 - \cos x}{1 + \cos x}\,dx = \int \left(\sec^2\dfrac{x}{2} - 1\right) dx$$

$$= 2\tan\dfrac{x}{2} - x + C$$

Write the numerator as $$\cos x = (1 + \cos x) - 1$$:

$$\dfrac{\cos x}{1 + \cos x} = 1 - \dfrac{1}{1 + \cos x}$$

Since $$1 + \cos x = 2\cos^2\dfrac{x}{2}$$, we have $$\dfrac{1}{1 + \cos x} = \dfrac{1}{2}\sec^2\dfrac{x}{2}$$.

Therefore

$$\displaystyle\int \dfrac{\cos x}{1 + \cos x}\,dx = \int \left(1 - \dfrac{1}{2}\sec^2\dfrac{x}{2}\right) dx$$

$$= x - \dfrac{1}{2}\cdot 2\tan\dfrac{x}{2} + C = x - \tan\dfrac{x}{2} + C$$

Write $$\sin^4 x = (\sin^2 x)^2 = \left(\dfrac{1 - \cos 2x}{2}\right)^2$$:

$$\sin^4 x = \dfrac{1}{4}\big(1 - 2\cos 2x + \cos^2 2x\big)$$

Now $$\cos^2 2x = \dfrac{1 + \cos 4x}{2}$$, so

$$\sin^4 x = \dfrac{1}{4}\left(1 - 2\cos 2x + \dfrac{1 + \cos 4x}{2}\right) = \dfrac{1}{4}\left(\dfrac{3}{2} - 2\cos 2x + \dfrac{\cos 4x}{2}\right)$$

Integrating:

$$\displaystyle\int \sin^4 x\,dx = \dfrac{1}{4}\left(\dfrac{3x}{2} - \sin 2x + \dfrac{\sin 4x}{8}\right) + C$$

$$= \dfrac{3x}{8} - \dfrac{\sin 2x}{4} + \dfrac{\sin 4x}{32} + C$$

Write $$\cos^4 2x = (\cos^2 2x)^2 = \left(\dfrac{1 + \cos 4x}{2}\right)^2$$:

$$\cos^4 2x = \dfrac{1}{4}\big(1 + 2\cos 4x + \cos^2 4x\big)$$

Now $$\cos^2 4x = \dfrac{1 + \cos 8x}{2}$$, so

$$\cos^4 2x = \dfrac{1}{4}\left(\dfrac{3}{2} + 2\cos 4x + \dfrac{\cos 8x}{2}\right)$$

Integrating:

$$\displaystyle\int \cos^4 2x\,dx = \dfrac{1}{4}\left(\dfrac{3x}{2} + \dfrac{2\sin 4x}{4} + \dfrac{\sin 8x}{16}\right) + C$$

$$= \dfrac{3x}{8} + \dfrac{\sin 4x}{8} + \dfrac{\sin 8x}{64} + C$$

Use $$\sin^2 x = 1 - \cos^2 x = (1 - \cos x)(1 + \cos x)$$:

$$\dfrac{\sin^2 x}{1 + \cos x} = \dfrac{(1 - \cos x)(1 + \cos x)}{1 + \cos x} = 1 - \cos x$$

Therefore

$$\displaystyle\int \dfrac{\sin^2 x}{1 + \cos x}\,dx = \int (1 - \cos x)\,dx = x - \sin x + C$$

Use $$\cos 2\theta = 2\cos^2\theta - 1$$ in the numerator:

$$\cos 2x - \cos 2\alpha = (2\cos^2 x - 1) - (2\cos^2\alpha - 1) = 2(\cos^2 x - \cos^2\alpha)$$

Factorise: $$\cos^2 x - \cos^2\alpha = (\cos x - \cos\alpha)(\cos x + \cos\alpha)$$, so

$$\dfrac{\cos 2x - \cos 2\alpha}{\cos x - \cos\alpha} = \dfrac{2(\cos x - \cos\alpha)(\cos x + \cos\alpha)}{\cos x - \cos\alpha} = 2(\cos x + \cos\alpha)$$

Therefore

$$\displaystyle\int \dfrac{\cos 2x - \cos 2\alpha}{\cos x - \cos\alpha}\,dx = 2\int (\cos x + \cos\alpha)\,dx$$

$$= 2\sin x + 2x\cos\alpha + C$$

Note that $$1 + \sin 2x = \sin^2 x + \cos^2 x + 2\sin x \cos x = (\sin x + \cos x)^2$$.

So the integrand is $$\dfrac{\cos x - \sin x}{(\sin x + \cos x)^2}$$.

Put $$t = \sin x + \cos x$$, so $$dt = (\cos x - \sin x)\,dx$$:

$$\displaystyle\int \dfrac{\cos x - \sin x}{(\sin x + \cos x)^2}\,dx = \int \dfrac{dt}{t^2} = -\dfrac{1}{t} + C$$

$$= -\dfrac{1}{\sin x + \cos x} + C$$

Write $$\tan^3 2x = \tan^2 2x \cdot \tan 2x = (\sec^2 2x - 1)\tan 2x$$, so

$$\tan^3 2x \sec 2x = (\sec^2 2x - 1)\,\sec 2x \tan 2x$$

Put $$t = \sec 2x$$, so $$dt = 2\sec 2x \tan 2x\,dx$$, i.e. $$\sec 2x \tan 2x\,dx = \dfrac{dt}{2}$$.

$$\displaystyle\int \tan^3 2x \sec 2x\,dx = \int (t^2 - 1)\dfrac{dt}{2} = \dfrac{1}{2}\left(\dfrac{t^3}{3} - t\right) + C$$

$$= \dfrac{\sec^3 2x}{6} - \dfrac{\sec 2x}{2} + C$$

Write $$\tan^4 x = \tan^2 x \cdot \tan^2 x = \tan^2 x\,(\sec^2 x - 1)$$:

$$\tan^4 x = \tan^2 x \sec^2 x - \tan^2 x = \tan^2 x \sec^2 x - (\sec^2 x - 1)$$

Integrate term by term. For $$\displaystyle\int \tan^2 x \sec^2 x\,dx$$ put $$t = \tan x$$, $$dt = \sec^2 x\,dx$$, giving $$\dfrac{t^3}{3} = \dfrac{\tan^3 x}{3}$$.

$$\displaystyle\int \tan^4 x\,dx = \dfrac{\tan^3 x}{3} - \int \sec^2 x\,dx + \int 1\,dx$$

$$= \dfrac{\tan^3 x}{3} - \tan x + x + C$$

Split into two fractions:

$$\dfrac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} = \dfrac{\sin^3 x}{\sin^2 x \cos^2 x} + \dfrac{\cos^3 x}{\sin^2 x \cos^2 x}$$

$$= \dfrac{\sin x}{\cos^2 x} + \dfrac{\cos x}{\sin^2 x} = \sec x \tan x + \operatorname{cosec} x \cot x$$

Therefore

$$\displaystyle\int \dfrac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x}\,dx = \int (\sec x \tan x + \operatorname{cosec} x \cot x)\,dx$$

$$= \sec x - \operatorname{cosec} x + C$$

Use $$\cos 2x = 1 - 2\sin^2 x$$ in the numerator:

$$\cos 2x + 2\sin^2 x = (1 - 2\sin^2 x) + 2\sin^2 x = 1$$

So the integrand simplifies to

$$\dfrac{\cos 2x + 2\sin^2 x}{\cos^2 x} = \dfrac{1}{\cos^2 x} = \sec^2 x$$

Therefore

$$\displaystyle\int \dfrac{\cos 2x + 2\sin^2 x}{\cos^2 x}\,dx = \int \sec^2 x\,dx = \tan x + C$$

Write the numerator $$1 = \sin^2 x + \cos^2 x$$:

$$\dfrac{1}{\sin x \cos^3 x} = \dfrac{\sin^2 x + \cos^2 x}{\sin x \cos^3 x} = \dfrac{\sin x}{\cos^3 x} + \dfrac{1}{\sin x \cos x}$$

For the first term, $$\dfrac{\sin x}{\cos^3 x} = \tan x \sec^2 x$$. Putting $$t = \tan x$$ gives $$\displaystyle\int \tan x \sec^2 x\,dx = \dfrac{\tan^2 x}{2}$$.

For the second term, $$\dfrac{1}{\sin x \cos x} = \dfrac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \tan x + \cot x$$, and

$$\displaystyle\int (\tan x + \cot x)\,dx = -\log |\cos x| + \log |\sin x| = \log |\tan x|$$

Combining:

$$\displaystyle\int \dfrac{dx}{\sin x \cos^3 x} = \dfrac{\tan^2 x}{2} + \log |\tan x| + C$$

Use $$\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$$:

$$\dfrac{\cos 2x}{(\cos x + \sin x)^2} = \dfrac{(\cos x - \sin x)(\cos x + \sin x)}{(\cos x + \sin x)^2} = \dfrac{\cos x - \sin x}{\cos x + \sin x}$$

Put $$t = \cos x + \sin x$$, so $$dt = (-\sin x + \cos x)\,dx = (\cos x - \sin x)\,dx$$:

$$\displaystyle\int \dfrac{\cos x - \sin x}{\cos x + \sin x}\,dx = \int \dfrac{dt}{t} = \log |t| + C$$

$$= \log |\cos x + \sin x| + C$$

Write $$\cos x = \sin\left(\dfrac{\pi}{2} - x\right)$$.

For $$x$$ in a range where $$\dfrac{\pi}{2} - x$$ lies in $$\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$ (i.e. $$0 \le x \le \pi$$),

$$\sin^{-1}(\cos x) = \sin^{-1}\left(\sin\left(\dfrac{\pi}{2} - x\right)\right) = \dfrac{\pi}{2} - x$$

Therefore

$$\displaystyle\int \sin^{-1}(\cos x)\,dx = \int \left(\dfrac{\pi}{2} - x\right) dx = \dfrac{\pi x}{2} - \dfrac{x^2}{2} + C$$

Multiply and divide by $$\sin(a - b)$$. Note that $$(x - b) - (x - a) = a - b$$, so

$$\sin(a - b) = \sin\big[(x - b) - (x - a)\big] = \sin(x - b)\cos(x - a) - \cos(x - b)\sin(x - a)$$

Therefore

$$\dfrac{1}{\cos(x - a)\cos(x - b)} = \dfrac{1}{\sin(a - b)} \cdot \dfrac{\sin(x - b)\cos(x - a) - \cos(x - b)\sin(x - a)}{\cos(x - a)\cos(x - b)}$$

$$= \dfrac{1}{\sin(a - b)}\big[\tan(x - b) - \tan(x - a)\big]$$

Integrating:

$$I = \dfrac{1}{\sin(a - b)}\big[-\log |\cos(x - b)| + \log |\cos(x - a)|\big] + C$$

$$= \dfrac{1}{\sin(a - b)}\log\left|\dfrac{\cos(x - a)}{\cos(x - b)}\right| + C$$

Split the fraction:

$$\dfrac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} = \dfrac{\sin^2 x}{\sin^2 x \cos^2 x} - \dfrac{\cos^2 x}{\sin^2 x \cos^2 x} = \dfrac{1}{\cos^2 x} - \dfrac{1}{\sin^2 x}$$

$$= \sec^2 x - \operatorname{cosec}^2 x$$

Therefore

$$\displaystyle\int (\sec^2 x - \operatorname{cosec}^2 x)\,dx = \tan x - (-\cot x) + C = \tan x + \cot x + C$$

This matches option (A).

Put $$t = x e^x$$. By the product rule,

$$dt = (e^x + x e^x)\,dx = e^x(1 + x)\,dx$$

So the integral becomes

$$\displaystyle\int \dfrac{e^x(1 + x)}{\cos^2(x e^x)}\,dx = \int \dfrac{dt}{\cos^2 t} = \int \sec^2 t\,dt$$

$$= \tan t + C = \tan(x e^x) + C$$

This matches option (B).

Let

$$I = \int \dfrac{3x^2}{x^6 + 1}\,dx$$

Step 1  (Substitution)
Choose $$t = x^3$$ because its derivative involves the factor $$3x^2$$ that appears in the integrand.

Then

$$dt = 3x^2\,dx \;\;\Longrightarrow\;\; 3x^2\,dx = dt$$

Step 2  (Rewrite the integral)

Substituting $$t$$ and $$dt$$ in the integral:

$$I = \int \dfrac{dt}{t^2 + 1}$$

because $$x^6 = (x^3)^2 = t^2$$.

Step 3  (Integrate)

The standard integral is

$$\int \dfrac{1}{1 + t^2}\,dt = \tan^{-1}t + C$$

Hence

$$I = \tan^{-1}t + C$$

Step 4  (Back-substitute)

Replace $$t$$ with $$x^3$$:

$$I = \tan^{-1}(x^3) + C$$

Therefore,

$$\displaystyle \int \dfrac{3x^2}{x^6 + 1}\,dx = \tan^{-1}(x^3) + C.$$

We have to evaluate the indefinite integral

$$\displaystyle \int \frac{1}{\sqrt{1+4x^2}}\,dx$$

Step 1 — Choose a trigonometric substitution.
Because the integrand contains the form $$1+4x^2$$ under a square root, put

$$2x = \tan\theta \;\;(\theta \text{ in radians}).$$

Then $$\tan\theta = 2x \implies \theta = \tan^{-1}(2x).$$

Differentiate both sides:

$$d\theta = \frac{2\,dx}{1+4x^2}\;\;\Longrightarrow\;\;dx = \frac{\sec^2\theta}{2}\,d\theta,$$

since $$1+\tan^2\theta = \sec^2\theta.$$

Step 2 — Rewrite the integrand.

First, compute the new denominator:

$$\sqrt{1+4x^2} = \sqrt{1+\tan^2\theta} = \sqrt{\sec^2\theta} = \sec\theta.$$

Now substitute $$x$$ and $$dx$$:

$$\int \frac{1}{\sqrt{1+4x^2}}\,dx = \int \frac{1}{\sec\theta}\,\left(\frac{\sec^2\theta}{2}\,d\theta\right) = \frac12 \int \sec\theta\,d\theta.$$

Step 3 — Integrate $$\sec\theta$$.

The standard result is

$$\int \sec\theta\,d\theta = \ln\left|\sec\theta+\tan\theta\right| + C.$$

Therefore,

$$\frac12 \int \sec\theta\,d\theta = \frac12\,\ln\left|\sec\theta+\tan\theta\right| + C.$$

Step 4 — Return to the variable $$x$$.

Recall the substitutions:

• $$\tan\theta = 2x,$$
• $$\sec\theta = \sqrt{1+\tan^2\theta}=\sqrt{1+4x^2}.$$

Replace $$\tan\theta$$ and $$\sec\theta$$ in the antiderivative:

$$\frac12\,\ln\left|\sec\theta+\tan\theta\right|+C = \frac12\,\ln\left|\sqrt{1+4x^2}+2x\right|+C.$$

Step 5 — Present the final result.

$$\boxed{\displaystyle\int \frac{dx}{\sqrt{1+4x^2}} = \frac12\,\ln\left|2x+\sqrt{1+4x^2}\right| + C}$$

Alternate form: Because $$\int\!\frac{du}{\sqrt{1+u^2}} = \sinh^{-1}u + C,$$ one may also write

$$\int \frac{dx}{\sqrt{1+4x^2}} = \frac12\,\sinh^{-1}(2x) + C,$$

since $$\sinh^{-1}(2x) = \ln\left|2x+\sqrt{1+4x^2}\right|.$$

Let the required integral be

$$I = \int \dfrac{dx}{\sqrt{(2 - x)^2 + 1}}.$$

1. Make a simple linear substitution.

Put $$u = 2 - x.$$ Then $$\dfrac{du}{dx} = -1 \;\;\Longrightarrow\;\; dx = -du.$$

2. Rewrite the integral in the new variable.

Substituting $$x$$ and $$dx$$ gives

$$I = \int \dfrac{-du}{\sqrt{u^2 + 1}} = -\int \dfrac{du}{\sqrt{u^2 + 1}}.$$

3. Recall the standard integral.

For any real constant $$a\,(\neq 0)$$,

$$\int \dfrac{dt}{\sqrt{t^2 + a^2}} = \ln\left|t + \sqrt{t^2 + a^2}\right| + C = \sinh^{-1}\!\left(\dfrac{t}{a}\right) + C.$$

With $$a = 1$$ and $$t = u$$ we have

$$\int \dfrac{du}{\sqrt{u^2 + 1}} = \sinh^{-1}(u) + C.$$

4. Integrate.

Therefore,

$$I = -\sinh^{-1}(u) + C.$$

5. Return to the original variable.

Recall that $$u = 2 - x$$, so

$$I = -\sinh^{-1}(2 - x) + C.$$

6. Simplify the final answer (optional).

The inverse hyperbolic sine function is odd: $$\sinh^{-1}(-z) = -\sinh^{-1}(z).$$ Hence

$$-\sinh^{-1}(2 - x) = \sinh^{-1}(x - 2).$$

Either of the following equivalent antiderivatives is acceptable:

• $$\boxed{\sinh^{-1}(x - 2) + C}$$
• $$\boxed{-\ln\left|2 - x + \sqrt{(2 - x)^2 + 1}\right| + C}$$

Thus the integral is completely evaluated.

Given integral:

$$I = \int \frac{dx}{\sqrt{9 - 25x^2}}$$

Step 1: Reduce to the standard form

Factor out $$9$$ from the square-root:

$$\sqrt{9 - 25x^2}=\sqrt{9\Bigl(1-\frac{25}{9}x^2\Bigr)} = 3\,\sqrt{1-\Bigl(\frac{5x}{3}\Bigr)^2}$$

Thus

$$I = \int \frac{dx}{3\,\sqrt{1-\bigl(\tfrac{5x}{3}\bigr)^2}} = \frac13\int \frac{dx}{\sqrt{1-\bigl(\tfrac{5x}{3}\bigr)^2}}$$

Step 2: Substitution

Put $$u = \dfrac{5x}{3} \;\Rightarrow\; x = \dfrac{3u}{5},\; dx = \dfrac{3}{5}\,du$$

Then

$$I = \frac13 \int \frac{\tfrac{3}{5}\,du}{\sqrt{1-u^2}} = \frac15 \int \frac{du}{\sqrt{1-u^2}}$$

Step 3: Integrate

The standard result $$\displaystyle \int \frac{du}{\sqrt{1-u^2}} = \sin^{-1}u + C$$ gives

$$I = \frac15\,\sin^{-1}u + C$$

Step 4: Back-substitute $$u = \dfrac{5x}{3}$$

$$I = \frac15\,\sin^{-1}\left(\frac{5x}{3}\right) + C$$

Hence the antiderivative of $$\dfrac{1}{\sqrt{9-25x^2}}$$ is $$\dfrac15\sin^{-1}\left(\dfrac{5x}{3}\right)+C$$.

Let the required integral be denoted by $$I$$:

$$I = \int \dfrac{3x}{1 + 2x^{4}}\,dx$$

Because the highest power of $$x$$ in the denominator is an even power, put $$t = x^{2}$$. Then

$$dt = 2x\,dx \;\;\Longrightarrow\;\; x\,dx = \dfrac{dt}{2}.$$

Substituting $$t$$ and $$dt$$ in $$I$$ gives

$$I = \int \dfrac{3x}{1 + 2x^{4}}\,dx = 3 \int \dfrac{x\,dx}{1 + 2(x^{2})^{2}} = 3 \int \dfrac{x\,dx}{1 + 2t^{2}}.$$

Replace $$x\,dx$$ by $$dt/2$$:

$$I = 3 \int \dfrac{\dfrac{dt}{2}}{1 + 2t^{2}} = \dfrac{3}{2}\int \dfrac{dt}{1 + 2t^{2}}.$$

Write the quadratic in the standard arctangent form:

$$1 + 2t^{2} = 1^{2} + (\sqrt{2}\,t)^{2}.$$

Using $$\displaystyle \int \dfrac{du}{a^{2}+u^{2}} = \dfrac{1}{a}\tan^{-1}\!\left(\dfrac{u}{a}\right)+C$$ with $$u = \sqrt{2}\,t$$ and $$a = 1$$, we have

$$\int \dfrac{dt}{1 + 2t^{2}} = \dfrac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}\,t) + C.$$

Therefore,

$$I = \dfrac{3}{2}\left( \dfrac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}\,t)\right) + C = \dfrac{3}{2\sqrt{2}}\tan^{-1}(\sqrt{2}\,t) + C.$$

Finally, re-substitute $$t = x^{2}$$:

$$I = \dfrac{3}{2\sqrt{2}}\tan^{-1}(\sqrt{2}\,x^{2}) + C.$$

Required integral: $$I=\displaystyle\int \frac{x^2}{1-x^6}\,dx$$

Step 1 : substitution $$u=x^3$$

Then $$du=3x^2\,dx\\ \Rightarrow x^2\,dx=\dfrac{du}{3}$$ and $$x^6=(x^3)^2=u^2$$.

Therefore

$$I=\int \frac{x^2}{1-x^6}\,dx=\int \frac{1}{3}\,\frac{du}{1-u^2}=\frac13\int\frac{du}{1-u^2}$$

Step 2 : resolve into partial fractions

Because $$1-u^2=(1-u)(1+u)$$ we look for constants $$A,B$$ such that

$$\frac1{1-u^2}=\frac{A}{1-u}+\frac{B}{1+u}.$$

Multiplying by $$(1-u)(1+u)$$ gives $$1=A(1+u)+B(1-u)=(A+B)+(A-B)u.$$

Equating coefficients:

• $$(A+B)=1$$
• $$(A-B)=0$$

Hence $$A=B=\dfrac12$$ and

$$\frac1{1-u^2}=\frac12\left(\frac1{1-u}+\frac1{1+u}\right).$$

Step 3 : integrate term by term

$$\frac13\int\frac{du}{1-u^2}=\frac13\cdot\frac12\int\left(\frac1{1-u}+\frac1{1+u}\right)du =\frac16\left[-\ln|1-u|+\ln|1+u|\right]+C$$

Combine the logarithms:

$$I=\frac16\ln\left|\frac{1+u}{1-u}\right|+C.$$

Step 4 : revert to $$x$$

Since $$u=x^3$$,

$$I=\frac16\ln\left|\frac{1+x^3}{1-x^3}\right|+C.$$

This is the required antiderivative.

We wish to evaluate the indefinite integral

$$\int \dfrac{x-1}{\sqrt{x^2-1}}\,dx.$$

First decompose the given fraction into two simpler parts:

$$\dfrac{x-1}{\sqrt{x^2-1}} = \dfrac{x}{\sqrt{x^2-1}} \,\; - \; \dfrac{1}{\sqrt{x^2-1}}.$$

Hence

$$\int \dfrac{x-1}{\sqrt{x^2-1}}\,dx = \int \dfrac{x}{\sqrt{x^2-1}}\,dx - \int \dfrac{1}{\sqrt{x^2-1}}\,dx.$$

Step 1. Evaluate $$\int \dfrac{x}{\sqrt{x^2-1}}\,dx$$.

Observe that

$$\dfrac{d}{dx}\Bigl(\sqrt{x^2-1}\Bigr) = \dfrac{x}{\sqrt{x^2-1}}.$$

Therefore the first integral is the direct antiderivative:

$$\int \dfrac{x}{\sqrt{x^2-1}}\,dx = \sqrt{x^2-1} + C_1.$$

Step 2. Evaluate $$\int \dfrac{1}{\sqrt{x^2-1}}\,dx$$.

Recall the standard form

$$\int \dfrac{dx}{\sqrt{x^2-a^2}} = \ln \left| x + \sqrt{x^2-a^2} \right| + C,$$

(valid for $$|x|>a$$). Setting $$a=1$$ gives

$$\int \dfrac{1}{\sqrt{x^2-1}}\,dx = \ln \left| x + \sqrt{x^2-1} \right| + C_2.$$

Step 3. Combine the results.

Putting the two parts together:

$$\int \dfrac{x-1}{\sqrt{x^2-1}}\,dx = \sqrt{x^2-1} - \ln \left| x + \sqrt{x^2-1} \right| + C,$$

where $$C = C_1 - C_2$$ is the overall constant of integration.

This completes the integration.

We have to evaluate the indefinite integral

$$I = \int \dfrac{x^2}{\sqrt{x^6 + a^6}}\,dx$$

Step 1  Reduce the power by putting $$t = x^3$$

Then $$dt = 3x^2\,dx \;\;\Longrightarrow\;\; x^2\,dx = \dfrac{dt}{3}$$

Substituting in the integral,

$$I = \int \dfrac{x^2}{\sqrt{x^6 + a^6}}\,dx = \dfrac13 \int \dfrac{dt}{\sqrt{t^2 + a^6}}$$

Step 2  Use the standard form

For any constant $$k>0$$,

$$\int \dfrac{du}{\sqrt{u^2 + k^2}} = \ln\left|u + \sqrt{u^2 + k^2}\right| + C$$

Comparing, we have $$u=t$$ and $$k=a^3$$. Hence

$$\dfrac13 \int \dfrac{dt}{\sqrt{t^2 + a^6}} = \dfrac13\,\ln\left|t + \sqrt{t^2 + a^6}\right| + C$$

Step 3  Return to the original variable

Replace $$t$$ by $$x^3$$:

$$I = \dfrac13\,\ln\left|x^3 + \sqrt{x^6 + a^6}\right| + C$$

Alternative form. Using $$\sinh^{-1}z = \ln\left(z + \sqrt{z^2 + 1}\right)$$, the answer can also be written as

$$I = \dfrac13\,\sinh^{-1}\left(\dfrac{x^3}{a^3}\right) + C$$

Let the required integral be

$$I = \int \dfrac{\sec^2 x}{\sqrt{\tan^2 x + 4}}\,dx.$$

The numerator $$\sec^2 x\,dx$$ reminds us of the derivative of $$\tan x$$, so we put

$$t = \tan x \;\;\;\Longrightarrow\;\;\; dt = \sec^2 x\,dx.$$

Substituting $$t$$ and $$dt$$ in the integral gives

$$I = \int \dfrac{dt}{\sqrt{t^2 + 4}}.$$

The quantity under the square-root is of the form $$t^2 + a^2$$ with $$a = 2$$. A standard integral from the table (or proved by differentiating) is

$$\int \dfrac{du}{\sqrt{u^2 + a^2}} = \ln\left|u + \sqrt{u^2 + a^2}\right| + C \; \;\bigl(\text{equivalently } \sinh^{-1}(u/a)+C\bigr).$$

Applying it with $$u \equiv t$$ and $$a = 2$$, we get

$$I = \ln\left|t + \sqrt{t^2 + 4}\right| + C.$$

Finally, revert to the original variable $$x$$ using $$t = \tan x$$:

$$I = \ln\left|\tan x + \sqrt{\tan^2 x + 4}\right| + C.$$

An equally acceptable equivalent form is

$$I = \sinh^{-1}\left(\dfrac{\tan x}{2}\right) + C,$$

because $$\sinh^{-1} y = \ln\left|y + \sqrt{y^2 + 1}\right|.$$

Thus the integral is completely evaluated.

We have to evaluate the integral

$$I = \displaystyle\int \dfrac{1}{\sqrt{x^2 + 2x + 2}}\,dx$$

Step 1: Complete the square in the denominator.

$$x^2 + 2x + 2 = (x+1)^2 + 1$$

Therefore

$$I = \displaystyle\int \dfrac{1}{\sqrt{(x+1)^2 + 1}}\,dx$$

Step 2: Make a substitution.

Let $$t = x + 1 \;\;\Rightarrow\;\; dt = dx$$.

With this, the integral becomes

$$I = \displaystyle\int \dfrac{1}{\sqrt{t^2 + 1}}\,dt$$

Step 3: Use the standard antiderivative.

For any real constant $$a$$, $$\displaystyle\int \dfrac{1}{\sqrt{t^2 + a^2}}\,dt = \ln \left| t + \sqrt{t^2 + a^2} \right| + C$$. Here $$a = 1$$, hence

$$I = \ln \left| t + \sqrt{t^2 + 1} \right| + C$$

Step 4: Back-substitute $$t = x + 1$$.

$$I = \ln \left| x + 1 + \sqrt{(x+1)^2 + 1} \right| + C$$

Since $$\sqrt{(x+1)^2 + 1} = \sqrt{x^2 + 2x + 2}$$, the required integral is

$$\boxed{\ln \left| x + 1 + \sqrt{x^2 + 2x + 2} \right| + C}$$

(An equivalent form is $$I = \sinh^{-1}(x+1) + C$$.)

We have to evaluate the integral

$$I = \int \dfrac{dx}{9x^2 + 6x + 5}$$

1. Complete the square in the denominator

$$9x^2 + 6x + 5 = 9\!\left(x^2 + \dfrac{2}{3}x\right) + 5$$

Inside the bracket, complete the square:

$$x^2 + \dfrac{2}{3}x = \left(x + \dfrac{1}{3}\right)^2 - \left(\dfrac{1}{3}\right)^2$$

Multiplying by 9:

$$9\left(x + \dfrac{1}{3}\right)^2 - 1$$

Now add the constant 5:

$$9\left(x + \dfrac{1}{3}\right)^2 - 1 + 5 = 9\left(x + \dfrac{1}{3}\right)^2 + 4$$

Thus

$$9x^2 + 6x + 5 = 9\left(x + \dfrac{1}{3}\right)^2 + 4$$

2. Introduce a substitution

Put $$u = x + \dfrac{1}{3} \;\;(\text{so } du = dx)$$ to obtain

$$I = \int \dfrac{du}{9u^2 + 4}$$

3. Factor the constant outside the integral

$$9u^2 + 4 = 9\left(u^2 + \dfrac{4}{9}\right) = 9\left(u^2 + \left(\dfrac{2}{3}\right)^2\right)$$

Therefore

$$I = \dfrac{1}{9} \int \dfrac{du}{u^2 + \left(\dfrac{2}{3}\right)^2}$$

4. Apply the standard formula

The identity $$\displaystyle \int \dfrac{du}{u^2 + a^2} = \dfrac{1}{a} \tan^{-1}\!\left(\dfrac{u}{a}\right) + C$$ holds for any constant $$a > 0$$.

Here $$a = \dfrac{2}{3}$$, so

$$\int \dfrac{du}{u^2 + \left(\dfrac{2}{3}\right)^2} = \dfrac{3}{2} \tan^{-1}\!\left(\dfrac{3u}{2}\right) + C$$

Add the prefactor $$\dfrac{1}{9}$$:

$$I = \dfrac{1}{9} \cdot \dfrac{3}{2} \tan^{-1}\!\left(\dfrac{3u}{2}\right) + C = \dfrac{1}{6} \tan^{-1}\!\left(\dfrac{3u}{2}\right) + C$$

5. Back-substitute $$u = x + \dfrac{1}{3}$$

Since $$3u = 3x + 1$$, we get

$$I = \dfrac{1}{6} \tan^{-1}\!\left(\dfrac{3x + 1}{2}\right) + C$$

Hence,

$$\displaystyle \int \dfrac{dx}{9x^2 + 6x + 5} = \dfrac{1}{6} \tan^{-1}\!\left(\dfrac{3x + 1}{2}\right) + C$$

We have to evaluate the integral

$$I = \displaystyle \int \frac{\mathrm d x}{\sqrt{7-6x-x^{2}}}.$$

1. Complete the square inside the square-root.

Rewrite the quadratic expression:

$$7-6x-x^{2}=-(x^{2}+6x-7).$$

Now complete the square for $$x^{2}+6x-7$$:

$$x^{2}+6x-7 = \bigl(x^{2}+6x+9\bigr)-9-7 = (x+3)^{2}-16.$$

Therefore,

$$7-6x-x^{2}= -\bigl((x+3)^{2}-16\bigr)=16-(x+3)^{2}.$$

So

$$I = \int \frac{\mathrm d x}{\sqrt{16-(x+3)^{2}}}.$$

2. Introduce a simple substitution.

Let

$$t = x+3 \;\;\Rightarrow\;\; \mathrm d t = \mathrm d x.$$

The integral becomes

$$I = \int \frac{\mathrm d t}{\sqrt{16-t^{2}}}.$$

3. Recognise the standard form.

The integral $$\displaystyle \int \frac{\mathrm d t}{\sqrt{a^{2}-t^{2}}}=\sin^{-1}\!\left(\tfrac{t}{a}\right)+C$$ is well known. Here $$a^{2}=16\;(a=4).$$ Hence

$$I = \sin^{-1}\!\left(\frac{t}{4}\right)+C.$$

4. Replace $$t$$ by $$x+3.$$

$$I = \sin^{-1}\!\left(\frac{x+3}{4}\right)+C.$$

Thus the required indefinite integral is

$$\boxed{\displaystyle \int \frac{\mathrm d x}{\sqrt{7-6x-x^{2}}}=\sin^{-1}\!\left(\frac{x+3}{4}\right)+C.}$$

We have to evaluate the indefinite integral

$$I = \int \dfrac{dx}{\sqrt{(x-1)(x-2)}}.$$

1. Expand the expression under the square–root

$$ (x-1)(x-2)=x^{2}-3x+2. $$

2. Complete the square

Rewrite the quadratic so that a perfect square appears:

$$x^{2}-3x+2=(x^{2}-3x+\tfrac94)+2-\tfrac94=(x-\tfrac32)^{2}-\tfrac14.$$

Therefore

$$ \sqrt{(x-1)(x-2)} = \sqrt{\,(x-\tfrac32)^{2}-\left(\tfrac12\right)^{2}\,}. $$

3. Translate the origin

Put

$$u = x-\tfrac32 \quad\Longrightarrow\quad du = dx.$$

In these terms

$$I = \int \dfrac{du}{\sqrt{u^{2}-\left(\tfrac12\right)^{2}}}. $$

4. Use the standard integral

For any positive constant $$a$$ we know

$$\int \dfrac{du}{\sqrt{u^{2}-a^{2}}}=\ln\left|u+\sqrt{u^{2}-a^{2}}\right|+C.$$

Taking $$a=\tfrac12$$ yields

$$I = \ln\left|u+\sqrt{u^{2}-\left(\tfrac12\right)^{2}}\right|+C.$$

5. Return to the variable $$x$$

Since $$u=x-\tfrac32,$$

$$I = \ln\left|\,x-\tfrac32+\sqrt{(x-1)(x-2)}\,\right|+C.$$

Thus, the required integral is obtained.

The integral to be evaluated is

$$I = \int \dfrac{\mathrm{d}x}{\sqrt{8+3x-x^{2}}}. $$

1. Complete the square in the quadratic expression.

Write the quadratic in the form $$a^{2}-u^{2}$$ by first factoring out the coefficient of $$x^{2}$$:

$$8+3x-x^{2}= -\bigl(x^{2}-3x-8\bigr).$$

Now complete the square inside the brackets:

\begin{align}x^{2}-3x-8 &= \left[x^{2}-3x+\left(\tfrac{3}{2}\right)^{2}\right]-\left(\tfrac{3}{2}\right)^{2}-8\\[4pt] &= \bigl(x-\tfrac{3}{2}\bigr)^{2}-\Bigl(\tfrac{9}{4}+8\Bigr)\\[4pt] &= \bigl(x-\tfrac{3}{2}\bigr)^{2}-\tfrac{41}{4}. \end{align}

Substituting this back with the minus sign:

$$8+3x-x^{2}=\frac{41}{4}-\bigl(x-\tfrac{3}{2}\bigr)^{2}.$$

2. Introduce a simple variable.

Put $$u=x-\dfrac{3}{2}\;\;(\Rightarrow\;\mathrm{d}u=\mathrm{d}x).$$ Then

$$I=\int\dfrac{\mathrm{d}u}{\sqrt{\dfrac{41}{4}-u^{2}}}.$$

3. Recognise the standard form.

Write the constant term as a square: $$\dfrac{41}{4}=\left(\dfrac{\sqrt{41}}{2}\right)^{2}.$$ Hence

$$I=\int\dfrac{\mathrm{d}u}{\sqrt{\left(\dfrac{\sqrt{41}}{2}\right)^{2}-u^{2}}}.$$

This matches the standard integral

$$\int\dfrac{\mathrm{d}x}{\sqrt{a^{2}-x^{2}}}=\sin^{-1}\!\left(\dfrac{x}{a}\right)+C.$$

With $$a=\dfrac{\sqrt{41}}{2}$$ and $$x\to u,$$ we have

$$I=\sin^{-1}\!\left(\dfrac{u}{\tfrac{\sqrt{41}}{2}}\right)+C.$$

4. Substitute back $$u=x-\dfrac{3}{2}$$.

\begin{align} I &= \sin^{-1}\!\left(\dfrac{x-\tfrac{3}{2}}{\tfrac{\sqrt{41}}{2}}\right)+C\\[6pt] &= \sin^{-1}\!\left(\dfrac{2x-3}{\sqrt{41}}\right)+C. \end{align}

Verification (optional): Differentiating $$\sin^{-1}\bigl((2x-3)/\sqrt{41}\bigr)$$ gives $$1/\sqrt{8+3x-x^{2}},$$ confirming the result.

Hence the required indefinite integral is found.

We wish to obtain an antiderivative of the function

$$I = \displaystyle \int \frac{\mathrm d x}{\sqrt{(x-a)(x-b)}} .$$

Step 1 – Preparation.
Without loss of generality assume $$b>a$$. (If $$a>b$$ simply interchange the symbols; the algebra that follows is identical.)
Because both roots are real, write the integral in a form that removes the parameters from the radical.

Step 2 – Linear substitution.
Put $$x=a+(b-a)u\;\;(u\in\mathbb R).$$
Then
$$x-a=(b-a)u,\qquad x-b=(b-a)(u-1),\qquad \mathrm dx=(b-a)\,\mathrm du.$$

Substituting these into $$I$$:

$$\begin{aligned} I &= \int \frac{(b-a)\,\mathrm du}{\sqrt{(b-a)u\;\bigl[(b-a)(u-1)\bigr]}} \\[2pt] &= \int \frac{(b-a)\,\mathrm du}{(b-a)\,\sqrt{u(u-1)}} \\[2pt] &= \int \frac{\mathrm du}{\sqrt{u(u-1)}} . \end{aligned}$$

The parameters $$a$$ and $$b$$ have disappeared; the remaining integral is a standard one.

Step 3 – Evaluating the reduced integral.
Let us show in detail that

$$\int \frac{\mathrm du}{\sqrt{u(u-1)}}=2\,\ln\left|\sqrt{u}+\sqrt{u-1}\right|+C.$$

Verification:
Set $$F(u)=2\ln\bigl|\sqrt{u}+\sqrt{u-1}\bigr|$$.
Then

$$\begin{aligned} F'(u)&=\frac{2}{\sqrt{u}+\sqrt{u-1}}\left(\frac{1}{2\sqrt{u}}+\frac{1}{2\sqrt{u-1}}\right)\\[4pt] &=\frac{1}{\sqrt{u}+\sqrt{u-1}}\cdot\frac{\sqrt{u-1}+\sqrt{u}}{\sqrt{u}\,\sqrt{u-1}}\\[4pt] &=\frac{1}{\sqrt{u}\,\sqrt{u-1}}=\frac{1}{\sqrt{u(u-1)}}. \end{aligned}$$

Hence the formula is correct. Therefore

$$I=2\ln\left|\sqrt{u}+\sqrt{u-1}\right|+C.$$

Step 4 – Re-substitution.
Recall $$u=\dfrac{x-a}{b-a}$$, so

$$\sqrt{u}=\frac{\sqrt{x-a}}{\sqrt{b-a}},\qquad \sqrt{u-1}=\frac{\sqrt{x-b}}{\sqrt{b-a}}.$$

Thus

$$\begin{aligned} I &=2\ln\left|\frac{\sqrt{x-a}+\sqrt{x-b}}{\sqrt{b-a}}\right|+C \\[4pt] &=2\ln\left|\sqrt{x-a}+\sqrt{x-b}\right|+C', \end{aligned}$$

where the constant $$-2\ln\sqrt{b-a}$$ has been absorbed into the new constant of integration $$C'$$.

Result.
The required integral is therefore

$$\boxed{\displaystyle \int \frac{\mathrm d x}{\sqrt{(x-a)(x-b)}}=2\ln\left|\sqrt{x-a}+\sqrt{x-b}\right|+C.}$$

(Any equivalent form, e.g. one involving a logarithm of a ratio, differs from the above only by an additive constant.)

We have to evaluate the integral

$$I = \int \frac{4x + 1}{\sqrt{2x^{2} + x - 3}}\,dx.$$

Step 1 – Choose a substitution.
Notice that the numerator $$4x + 1$$ is exactly the derivative of the quadratic expression in the square-root denominator:

$$\frac{d}{dx}\left(2x^{2} + x - 3\right) = 4x + 1.$$

This suggests the substitution

$$t = 2x^{2} + x - 3.$$

Step 2 – Compute $$dt$$.

$$dt = (4x + 1)\,dx \quad\Longrightarrow\quad dx = \frac{dt}{4x + 1}.$$

Step 3 – Rewrite the integral.

Substituting $$t$$ and $$dt$$ into $$I$$, we get

$$I = \int \frac{4x + 1}{\sqrt{t}}\;\frac{dt}{4x + 1} = \int \frac{dt}{\sqrt{t}}.$$

Step 4 – Integrate in $$t$$.

Recall that $$\int t^{-1/2}\,dt = 2t^{1/2} + C.$$ Hence

$$\int \frac{dt}{\sqrt{t}} = 2\sqrt{t} + C.$$

Step 5 – Back-substitute $$t = 2x^{2} + x - 3$$.

Therefore,

$$I = 2\sqrt{2x^{2} + x - 3} + C.$$

Final result

$$\boxed{\displaystyle \int \frac{4x + 1}{\sqrt{2x^{2} + x - 3}}\,dx = 2\sqrt{2x^{2} + x - 3} + C.}$$

We have to evaluate the indefinite integral

$$I = \int \dfrac{x + 2}{\sqrt{x^2 - 1}}\,dx.$$

First separate the numerator:

$$I = \int \left( \dfrac{x}{\sqrt{x^2 - 1}} \, + \, \dfrac{2}{\sqrt{x^2 - 1}} \right) dx = \int \dfrac{x}{\sqrt{x^2 - 1}}\,dx \; + \; 2 \int \dfrac{dx}{\sqrt{x^2 - 1}}.$$

1. Integrating $$\displaystyle \int \dfrac{x}{\sqrt{x^2 - 1}}\,dx$$

Put $$u = x^2 - 1 \;\Rightarrow\; du = 2x\,dx \;\Rightarrow\; \dfrac{du}{2} = x\,dx.$$

Substituting,

$$\int \dfrac{x}{\sqrt{x^2 - 1}}\,dx = \int \dfrac{\tfrac{du}{2}}{\sqrt{u}} = \tfrac12 \int u^{-1/2}\,du.$$

Integrate:

$$\tfrac12 \cdot \dfrac{u^{1/2}}{\tfrac12} = u^{1/2} = \sqrt{u}.$$

Restore $$u = x^2 - 1$$:

$$\int \dfrac{x}{\sqrt{x^2 - 1}}\,dx = \sqrt{x^2 - 1} + C_1.$$

2. Integrating $$\displaystyle 2 \int \dfrac{dx}{\sqrt{x^2 - 1}}$$

The standard result

$$\int \dfrac{dx}{\sqrt{x^2 - a^2}} = \ln\left|x + \sqrt{x^2 - a^2}\right| + C,$$

applied with $$a = 1$$ gives

$$\int \dfrac{dx}{\sqrt{x^2 - 1}} = \ln\left|x + \sqrt{x^2 - 1}\right| + C_2.$$

Multiplying by 2:

$$2 \int \dfrac{dx}{\sqrt{x^2 - 1}} = 2\ln\left|x + \sqrt{x^2 - 1}\right| + C_3.$$

3. Combine the results

Adding the two parts (and absorbing all constants into a single constant $$C$$):

$$I = \sqrt{x^2 - 1} + 2\ln\left|x + \sqrt{x^2 - 1}\right| + C.$$

Hence the required integral is

$$\boxed{\displaystyle \int \dfrac{x + 2}{\sqrt{x^2 - 1}}\,dx = \sqrt{x^2 - 1} + 2\ln\left|x + \sqrt{x^2 - 1}\right| + C.}$$

Given integral

$$I = \int \dfrac{5x-2}{1+2x+3x^2}\,dx$$

Step 1 — Express the numerator in terms of the derivative of the denominator

The denominator is $$D(x)=1+2x+3x^2$$, so $$D'(x)=2+6x=6x+2$$.

We look for constants $$A,B$$ such that

$$5x-2=A(6x+2)+B.$$

Equating coefficients of like powers of $$x$$ gives

• For $$x^1$$: $$5=6A\;\Rightarrow\;A=\dfrac56.$$
• For $$x^0$$: $$-2=2A+B\;\Rightarrow\;B=-2-\dfrac{10}{6}=-\dfrac{11}{3}.$$

Hence

$$5x-2=\dfrac56\,(6x+2)-\dfrac{11}{3}.$$

Step 2 — Split the integral

$$I=\int \left[\dfrac56\,\dfrac{6x+2}{D(x)}-\dfrac{11}{3}\,\dfrac{1}{D(x)}\right]dx =\dfrac56\int \dfrac{D'(x)}{D(x)}\,dx-\dfrac{11}{3}\int \dfrac{dx}{1+2x+3x^2}.$$

Step 3 — The logarithmic part

$$\dfrac56\int \dfrac{D'(x)}{D(x)}dx=\dfrac56\,\ln|D(x)|+C_1 =\dfrac56\,\ln|1+2x+3x^2|+C_1.$$

Step 4 — The quadratic part

To integrate $$\displaystyle\int \dfrac{dx}{1+2x+3x^2}$$ we first complete the square in the denominator.

$$1+2x+3x^2=3x^2+2x+1=3\Bigl(x^2+\dfrac23x\Bigr)+1.$$

Inside the brackets:

$$x^2+\dfrac23x=(x+\dfrac13)^2-\dfrac19,$$

so

$$1+2x+3x^2=3\Bigl[(x+\tfrac13)^2-\tfrac19\Bigr]+1 =3(x+\tfrac13)^2-\tfrac13+1=3(x+\tfrac13)^2+\tfrac23.$$

Factor the $$3$$:

$$1+2x+3x^2=3\Bigl[(x+\tfrac13)^2+\tfrac29\Bigr].$$

Therefore

$$\int \dfrac{dx}{1+2x+3x^2}=\dfrac13\int \dfrac{dx}{(x+\tfrac13)^2+\tfrac29}.$$

Let $$u=x+\dfrac13\;\Rightarrow\;du=dx.$$ The integral becomes

$$\dfrac13\int \dfrac{du}{u^2+\bigl(\sqrt{\tfrac29}\bigr)^2}.$$

Since $$\sqrt{\tfrac29}=\dfrac{\sqrt2}{3},$$ set $$a=\dfrac{\sqrt2}{3}.$$ Using $$\displaystyle\int \dfrac{du}{u^2+a^2}=\dfrac1a\arctan\!\left(\dfrac{u}{a}\right)+C,$$ we get

$$\int \dfrac{dx}{1+2x+3x^2}=\dfrac13\cdot\dfrac{1}{a}\,\arctan\!\left(\dfrac{u}{a}\right)+C_2 =\dfrac13\cdot\dfrac{3}{\sqrt2}\,\arctan\!\left(\dfrac{x+\tfrac13}{\tfrac{\sqrt2}{3}}\right)+C_2 =\dfrac1{\sqrt2}\,\arctan\!\left(\dfrac{3x+1}{\sqrt2}\right)+C_2.$$

Step 5 — Combine both parts

Bringing back the factor $$-\dfrac{11}{3}$$ from Step 2:

$$-\dfrac{11}{3}\int \dfrac{dx}{1+2x+3x^2}=-\dfrac{11}{3}\cdot\dfrac1{\sqrt2}\, \arctan\!\left(\dfrac{3x+1}{\sqrt2}\right) =-\dfrac{11}{3\sqrt2}\,\arctan\!\left(\dfrac{3x+1}{\sqrt2}\right).$$

Step 6 — Final result

Combining all constant multiples into a single constant $$C$$, the required integral is

$$\boxed{\displaystyle\int \dfrac{5x-2}{1+2x+3x^2}\,dx =\dfrac56\,\ln|1+2x+3x^2|\;- \dfrac{11}{3\sqrt2}\,\arctan\!\left(\dfrac{3x+1}{\sqrt2}\right)+C.}$$

Objective  Integrate

$$I= \int \dfrac{6x+7}{\sqrt{(x-5)(x-4)}}\,dx.$$

1. Express the quadratic under the root in a convenient form

$$(x-5)(x-4)=x^{2}-9x+20.$$

The derivative of this quadratic is

$$\dfrac{d}{dx}\bigl(x^{2}-9x+20\bigr)=2x-9.$$

2. Split the numerator so that part of it is a multiple of $$2x-9$$

Find constants $$A,B$$ such that

$$6x+7=A(2x-9)+B.$$

Comparing coefficients:

• Coefficient of $$x$$:  $$6=2A\;\Rightarrow\;A=3$$
• Constant term:  $$7=-9A+B\;\Rightarrow\;B=7+9\cdot3=34$$

Hence

$$6x+7=3(2x-9)+34.$$

3. Rewrite the integral with this decomposition

$$I= \int\dfrac{3(2x-9)+34}{\sqrt{(x-5)(x-4)}}\,dx =3\int\dfrac{2x-9}{\sqrt{(x-5)(x-4)}}\,dx+34\int\dfrac{dx}{\sqrt{(x-5)(x-4)}}.$$

4. First integral — a direct substitution

Let $$u=(x-5)(x-4)=x^{2}-9x+20.$$ Then $$du=(2x-9)\,dx.$$

Therefore

$$\int\dfrac{2x-9}{\sqrt{(x-5)(x-4)}}\,dx=\int\dfrac{du}{\sqrt u}=2\sqrt u=2\sqrt{(x-5)(x-4)}.$$

Multiplying by the factor $$3$$ already outside gives

$$3\times2\sqrt{(x-5)(x-4)}=6\sqrt{(x-5)(x-4)}.$$

5. Second integral — complete the square

Write the quadratic inside the root as a difference of two squares:

$$x^{2}-9x+20=(x-\tfrac92)^2-\bigl(\tfrac12\bigr)^2.$$

Thus

$$\int\dfrac{dx}{\sqrt{(x-5)(x-4)}}= \int\dfrac{dx}{\sqrt{(x-\tfrac92)^2-(\tfrac12)^2}}.$$

For any constants $$a,b>0,$$ the standard result is

$$\int\dfrac{dx}{\sqrt{(x-a)^2-b^2}}= \ln\left|\,x-a+\sqrt{(x-a)^2-b^2}\,\right|+C.$$

Here $$a=\dfrac92,\ b=\dfrac12,$$ so

$$\int\dfrac{dx}{\sqrt{(x-5)(x-4)}}= \ln\left|\,(x-\tfrac92)+\sqrt{(x-5)(x-4)}\,\right|+C.$$

Multiplying by the constant $$34$$ gives

$$34\ln\left|\,(x-\tfrac92)+\sqrt{(x-5)(x-4)}\,\right|.$$

6. Combine both parts and add the constant of integration

$$I=6\sqrt{(x-5)(x-4)}+34\ln\left|\,(x-\tfrac92)+\sqrt{(x-5)(x-4)}\,\right|+C.$$

Therefore

$$\boxed{\displaystyle\int\dfrac{6x+7}{\sqrt{(x-5)(x-4)}}\,dx=6\sqrt{(x-5)(x-4)}+34\ln\left|\,x-\tfrac92+\sqrt{(x-5)(x-4)}\,\right|+C.}$$

We have to evaluate the integral

$$I=\int \dfrac{x+2}{\sqrt{4x-x^{2}}}\,dx.$$

1. Make the quadratic in the square root a perfect square.
   

   $$4x-x^{2}=-(x^{2}-4x)=-(x^{2}-4x+4-4)=4-(x-2)^{2}.$$
   Thus $$\sqrt{4x-x^{2}}=\sqrt{4-(x-2)^{2}}.$$

2. Trigonometric substitution.
   

   Put $$x-2=2\sin\theta\;\;(\,-\pi/2\le \theta\le\pi/2\,)\;\;\Longrightarrow\;\;x=2+2\sin\theta.$$
   Then $$dx=2\cos\theta\,d\theta,$$ and

   $$\sqrt{4-(x-2)^{2}}=\sqrt{4-4\sin^{2}\theta}=2\cos\theta.$$

3. Transform the integrand.
   

   The numerator becomes $$x+2=(2+2\sin\theta)+2=4+2\sin\theta.$$

   Therefore

   $$I=\int \dfrac{4+2\sin\theta}{2\cos\theta}\,(2\cos\theta\,d\theta)=\int(4+2\sin\theta)\,d\theta.$$

4. Integrate in \(\theta\).
   

   $$\int(4+2\sin\theta)\,d\theta=4\theta-2\cos\theta+C.$$

5. Return to the variable \(x\).
   

   From the substitution $$\sin\theta=\dfrac{x-2}{2}\;\;\Longrightarrow\;\;\theta=\sin^{-1}\Bigl(\dfrac{x-2}{2}\Bigr).$$

   Also $$\cos\theta=\sqrt{1-\sin^{2}\theta}=\sqrt{1-\dfrac{(x-2)^{2}}{4}}=\dfrac{\sqrt{4-(x-2)^{2}}}{2}=\dfrac{\sqrt{4x-x^{2}}}{2}.$$

   Hence $$-2\cos\theta=-\sqrt{4x-x^{2}}.$$

   Substituting back,

   $$I=4\sin^{-1}\Bigl(\dfrac{x-2}{2}\Bigr)-\sqrt{4x-x^{2}}+C.$$

Thus the required integral is obtained.

We need to evaluate the indefinite integral

$$\displaystyle\;I \,=\,\int \dfrac{x+2}{\sqrt{x^{2}+2x+3}}\,dx$$

1. Complete the square in the quadratic.

$$x^{2}+2x+3\;=\;(x+1)^{2}+2$$

Hence write the integrand in terms of the new variable

$$t\;=\;x+1\;\;\Longrightarrow\;\;dt\;=\;dx$$

Express the numerator in t:

$$x+2\;=\;(t-1)+2\;=\;t+1$$

Therefore

$$I\;=\;\int \dfrac{t+1}{\sqrt{t^{2}+2}}\,dt$$

2. Split the integral.

$$I\;=\;\int \dfrac{t}{\sqrt{t^{2}+2}}\,dt\;\; +\;\; \int \dfrac{1}{\sqrt{t^{2}+2}}\,dt$$

Denote the two parts by $$I_{1}$$ and $$I_{2}$$, respectively.

3. Evaluate $$I_{1} = \displaystyle\int \dfrac{t}{\sqrt{t^{2}+2}}\,dt$$.

Put $$u = t^{2}+2 \;\;\Longrightarrow\;\;du = 2t\,dt \;\;\Longrightarrow\;\;t\,dt = \dfrac{du}{2}$$.

Then

$$I_{1}=\int \dfrac{t}{\sqrt{t^{2}+2}}\,dt\;=\;\int \dfrac{1}{2}\,u^{-\tfrac12}\,du\;=\;\dfrac12\cdot \dfrac{u^{\tfrac12}}{\tfrac12}\;=\;\sqrt{u}\;=\;\sqrt{t^{2}+2}$$

4. Evaluate $$I_{2} = \displaystyle\int \dfrac{1}{\sqrt{t^{2}+2}}\,dt$$.

This is the standard form $$\int \dfrac{1}{\sqrt{t^{2}+a^{2}}}\,dt = \ln\left| t+\sqrt{t^{2}+a^{2}} \right| + C$$ with $$a^{2}=2\;(a=\sqrt2)$$.

Hence

$$I_{2}=\ln\left| t+\sqrt{t^{2}+2} \right|$$

5. Combine the two parts and revert to $$x$$.

$$I\;=\;\sqrt{t^{2}+2}\; +\; \ln\left| t+\sqrt{t^{2}+2} \right|\; +\; C$$

Restore $$t=x+1$$:

$$I\;=\;\sqrt{(x+1)^{2}+2}\; +\; \ln\left|\,x+1 + \sqrt{(x+1)^{2}+2}\,\right|\; +\; C$$

Finally, replace the square again:

$$I\;=\;\sqrt{x^{2}+2x+3}\; +\; \ln\left|\,x+1 + \sqrt{x^{2}+2x+3}\,\right|\; +\; C$$

This completes the integration.

We have to evaluate

$$I = \int \dfrac{x+3}{x^{2}-2x-5}\,dx.$$

1. Express the numerator in terms of the derivative of the denominator.

Let $$D = x^{2}-2x-5 \;\Rightarrow\; D' = 2x-2 = 2(x-1).$$

Seek constants A and B such that

$$x+3 = A\,D' + B = A(2x-2)+B.$$

Comparing coefficients,

• Coefficient of $$x$$: $$1 = 2A \;\Rightarrow\; A = \dfrac12.$$
• Constant term: $$3 = -2A + B \;\Rightarrow\; B = 3 + 2A = 3 + 1 = 4.$$

Hence

$$x+3 = \dfrac12(2x-2) + 4 = \dfrac12 D' + 4.$$

2. Split the integral.

$$I = \int \dfrac{\frac12 D'}{D}\,dx + \int \dfrac{4}{x^{2}-2x-5}\,dx = \dfrac12\int \dfrac{D'}{D}\,dx + 4\int \dfrac{dx}{x^{2}-2x-5}.$$

3. Integrate the first part.

$$\dfrac12\int \dfrac{D'}{D}\,dx = \dfrac12\ln|D| + C_{1} = \dfrac12\ln|x^{2}-2x-5| + C_{1}.$$

4. Integrate the second part.

First complete the square in the denominator:

$$x^{2}-2x-5 = (x-1)^{2}-6.$$

Let $$u = x-1 \;\Rightarrow\; du = dx.$$ Then

$$\int \dfrac{dx}{x^{2}-2x-5} = \int \dfrac{du}{u^{2}-6}.$$

Writing $$6 = (\sqrt6)^{2}$$, the standard formula

$$\int \dfrac{du}{u^{2}-a^{2}} = \dfrac{1}{2a}\ln\left|\dfrac{u-a}{u+a}\right| + C$$

gives

$$\int \dfrac{du}{u^{2}-6} = \dfrac{1}{2\sqrt6} \ln\left|\dfrac{u-\sqrt6}{u+\sqrt6}\right| + C_{2}.$$ Substituting $$u = x-1$$,

$$\int \dfrac{dx}{x^{2}-2x-5} = \dfrac{1}{2\sqrt6} \ln\left|\dfrac{x-1-\sqrt6}{x-1+\sqrt6}\right| + C_{2}.$$

Multiplying by 4,

$$4\int \dfrac{dx}{x^{2}-2x-5} = \dfrac{2}{\sqrt6} \ln\left|\dfrac{x-1-\sqrt6}{x-1+\sqrt6}\right| + C_{3}.$$

5. Combine the results.

Absorbing all constants into a single $$C$$, we obtain

$$I = \dfrac12\ln|x^{2}-2x-5| + \dfrac{2}{\sqrt6}\ln\left|\dfrac{x-1-\sqrt6}{x-1+\sqrt6}\right| + C.$$

Thus the required integral is

$$\boxed{\displaystyle\int \frac{x+3}{x^{2}-2x-5}\,dx = \frac12\ln|x^{2}-2x-5| + \frac{2}{\sqrt6} \ln\left|\frac{x-1-\sqrt6}{x-1+\sqrt6}\right| + C}.$$

Let

$$I = \int \dfrac{5x + 3}{\sqrt{x^{2} + 4x + 10}}\,dx.$$

First complete the square in the quadratic that appears in the denominator:

$$x^{2} + 4x + 10 = (x + 2)^{2} + 6.$$

We next try to express the numerator $$5x + 3$$ in the form $$A\,(2x + 4) + B$$, because $$2x + 4$$ is the derivative of $$x^{2} + 4x + 10$$.

Equating coefficients,

$$5x + 3 \equiv A(2x + 4) + B = 2A\,x + (4A + B).$$

Hence

• Coefficient of $$x$$: $$2A = 5 \;\Longrightarrow\; A = \dfrac{5}{2}.$$
• Constant term: $$4A + B = 3 \;\Longrightarrow\; 4\left(\dfrac{5}{2}\right) + B = 3 \;\Longrightarrow\; 10 + B = 3 \;\Longrightarrow\; B = -7.$$

Therefore

$$5x + 3 = \frac{5}{2}(2x + 4) - 7.$$

Substituting this back,

$$I = \int \dfrac{\frac{5}{2}(2x + 4) - 7}{\sqrt{x^{2} + 4x + 10}}\,dx = \frac{5}{2}\int \dfrac{2x + 4}{\sqrt{x^{2} + 4x + 10}}\,dx - 7\int \dfrac{dx}{\sqrt{x^{2} + 4x + 10}}.$$

First integral

Put $$u = x^{2} + 4x + 10 \;\Rightarrow\; du = (2x + 4)\,dx.$$

Then

$$\frac{5}{2}\int \dfrac{2x + 4}{\sqrt{x^{2} + 4x + 10}}\,dx = \frac{5}{2}\int \dfrac{du}{\sqrt{u}} = \frac{5}{2}\cdot 2\sqrt{u} = 5\sqrt{u} = 5\sqrt{x^{2} + 4x + 10}.$$

Second integral

Write again $$x^{2} + 4x + 10 = (x+2)^{2} + 6.$$ Let $$t = x + 2 \;\Rightarrow\; dt = dx.$$ Then

$$\int \dfrac{dx}{\sqrt{x^{2} + 4x + 10}} = \int \dfrac{dt}{\sqrt{t^{2} + 6}}.$$

A standard result is

$$\int \dfrac{dt}{\sqrt{t^{2} + a^{2}}} = \ln\left| t + \sqrt{t^{2} + a^{2}} \right| + C.$$

With $$a^{2} = 6$$ we obtain

$$\int \dfrac{dt}{\sqrt{t^{2} + 6}} = \ln\left| t + \sqrt{t^{2} + 6} \right| + C.$$

Returning to $$x$$,

$$\int \dfrac{dx}{\sqrt{x^{2} + 4x + 10}} = \ln\left| x + 2 + \sqrt{x^{2} + 4x + 10} \right| + C.$$

Multiplying by the factor $$-7$$, the second integral contributes

$$-7\ln\left| x + 2 + \sqrt{x^{2} + 4x + 10} \right|.$$

Combining both parts

$$I = 5\sqrt{x^{2} + 4x + 10} - 7\ln\left| x + 2 + \sqrt{x^{2} + 4x + 10} \right| + C,$$

where $$C$$ is the constant of integration.

We have to evaluate

$$\displaystyle\int \frac{dx}{x^{2}+2x+2}$$

Step 1 — Complete the square in the denominator.

$$x^{2}+2x+2 = \bigl(x^{2}+2x+1\bigr)+1 = (x+1)^{2}+1$$

Step 2 — Rewrite the integral.

$$\int \frac{dx}{(x+1)^{2}+1}$$

Step 3 — Substitute.

Put $$t = x+1 \;\Longrightarrow\; dt = dx$$.

The integral becomes

$$\int \frac{dt}{t^{2}+1}$$

Step 4 — Recall the standard form.

For all real $$t$$,

$$\int \frac{dt}{t^{2}+1} = \tan^{-1} t + C$$

Hence,

$$\int \frac{dx}{x^{2}+2x+2} = \tan^{-1}(x+1) + C$$

Step 5 — Match with the given options.

The result coincides with option (B).

Let

$$I = \int \frac{dx}{\sqrt{9x-4x^{2}}}$$

1. Complete the square inside the root

Rewrite the quadratic:

$$9x-4x^{2}= -4x^{2}+9x$$

Factor −4 and complete the square in the bracket:

$$-4x^{2}+9x = -4\Bigl(x^{2}-\tfrac{9}{4}x\Bigr)$$

Add and subtract $$\bigl(\tfrac{9}{8}\bigr)^{2}$$ inside:

$$x^{2}-\tfrac{9}{4}x = \Bigl(x-\tfrac{9}{8}\Bigr)^{2}-\Bigl(\tfrac{9}{8}\Bigr)^{2}$$

Hence

$$9x-4x^{2}= -4\Bigl[\Bigl(x-\tfrac{9}{8}\Bigr)^{2}-\Bigl(\tfrac{9}{8}\Bigr)^{2}\Bigr]=4\Bigl[\Bigl(\tfrac{9}{8}\Bigr)^{2}-\Bigl(x-\tfrac{9}{8}\Bigr)^{2}\Bigr].$$

2. Express the integrand in standard form

Therefore

$$\sqrt{9x-4x^{2}} = 2\sqrt{\Bigl(\tfrac{9}{8}\Bigr)^{2}-\Bigl(x-\tfrac{9}{8}\Bigr)^{2}}.$$

So

$$I = \int \frac{dx}{2\sqrt{\bigl(\tfrac{9}{8}\bigr)^{2}-(x-\tfrac{9}{8})^{2}}} = \tfrac12\int \frac{dx}{\sqrt{\bigl(\tfrac{9}{8}\bigr)^{2}-(x-\tfrac{9}{8})^{2}}}.$$

3. Substitution

Put $$u=x-\tfrac{9}{8}\;(\Rightarrow\;du=dx).$$ Then

$$I = \tfrac12\int \frac{du}{\sqrt{\bigl(\tfrac{9}{8}\bigr)^{2}-u^{2}}}.$$

This is the standard integral $$\int \frac{du}{\sqrt{a^{2}-u^{2}}}=\sin^{-1}\!\left(\tfrac{u}{a}\right)+C.$$ Here $$a=\tfrac{9}{8}.$$

Hence

$$I = \tfrac12\sin^{-1}\!\left(\frac{u}{9/8}\right)+C = \tfrac12\sin^{-1}\!\left(\frac{8u}{9}\right)+C.$$

Re-substitute $$u=x-\tfrac{9}{8}$$:

$$I = \tfrac12\sin^{-1}\!\left(\frac{8\bigl(x-\tfrac{9}{8}\bigr)}{9}\right)+C = \tfrac12\sin^{-1}\!\left(\frac{8x-9}{9}\right)+C.$$

4. Choose the correct option

This matches option (B).

We resolve the integrand into partial fractions.

Assume $$\frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$$ with constants A and B.

Multiply by $$(x+1)(x+2)$$:

$$1 = A(x+2) + B(x+1).$$

Choose convenient values of $$x$$ to determine A and B.

• Put $$x = -2$$: $$1 = B(-2+1) = -B \;\;\Rightarrow\;\; B = -1.$$
• Put $$x = -1$$: $$1 = A(-1+2) = A \;\;\Rightarrow\;\; A = 1.$$

Hence $$\frac{1}{(x+1)(x+2)} = \frac{1}{x+1} - \frac{1}{x+2}.$$

Integrate term by term:

$$\int \frac{dx}{(x+1)(x+2)} = \int \frac{dx}{x+1} - \int \frac{dx}{x+2}.$$

Each integral is elementary:

$$\int \frac{dx}{x+1} = \ln|x+1| + C_1, \qquad \int \frac{dx}{x+2} = \ln|x+2| + C_2.$$

Merging the constants (write $$C = C_1 - C_2$$):

$$\int \frac{dx}{(x+1)(x+2)} = \ln|x+1| - \ln|x+2| + C.$$

Using $$\ln a - \ln b = \ln\frac{a}{b}$$ we obtain the compact form

$$\boxed{\displaystyle \int \frac{dx}{(x+1)(x+2)} = \ln\left|\frac{x+1}{x+2}\right| + C}. $$

Integral to be evaluated
$$I = \int \dfrac{x^2 + 1}{x^2 - 5x + 6}\,dx$$

1. Factor the denominator
$$x^2 - 5x + 6 = (x-2)(x-3).$$

2. Perform polynomial division
Write the numerator as a multiple of the denominator plus a remainder:

$$x^2 + 1 = \bigl(x^2 - 5x + 6\bigr) + (5x - 5).$$

Hence

$$\frac{x^2 + 1}{x^2 - 5x + 6} = 1 + \frac{5x - 5}{x^2 - 5x + 6}.$$

So

$$I = \int 1\,dx + \int \frac{5x - 5}{x^2 - 5x + 6}\,dx.$$

3. Decompose the second integral

The derivative of the denominator is $$\dfrac{d}{dx}(x^2 - 5x + 6)=2x-5.$$ Express the numerator $$5x-5$$ in terms of $$2x-5$$:

Find constants $$k,c$$ such that $$5x-5 = k(2x-5)+c.$$

Comparing coefficients:
For $$x$$: $$5 = 2k \Rightarrow k = \tfrac52.$$
Constant term: $$-5 = -5k + c \Rightarrow c = -5 + 5k = -5 + \tfrac{25}{2} = \tfrac{15}{2}.$$

Thus

$$5x - 5 = \tfrac52\,(2x-5) + \tfrac{15}{2}.$$

Therefore

$$\frac{5x - 5}{x^2 - 5x + 6} = \tfrac52\,\frac{2x-5}{x^2 - 5x + 6} + \tfrac{15}{2}\,\frac{1}{x^2 - 5x + 6}.$$

Split the integral accordingly:

$$I = \int 1\,dx + \tfrac52 \int \frac{2x-5}{x^2 - 5x + 6}\,dx + \tfrac{15}{2}\int \frac{1}{(x-2)(x-3)}\,dx.$$

4. Evaluate each part

• First part: $$\int 1\,dx = x.$$
• Second part: Since the numerator is the derivative of the denominator,
  $$\int \frac{2x-5}{x^2 - 5x + 6}\,dx = \ln|x^2 - 5x + 6|.$$ Therefore it contributes $$\tfrac52\,\ln|x^2 - 5x + 6|.$$
• Third part: Use partial fractions.
  Write
  $$\frac{1}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}.$$
  Equating numerators:
  $$A(x-3) + B(x-2) = 1.$$
  Setting $$x=3$$ gives $$B=1,$$ and setting $$x=2$$ gives $$A=-1.$$
  Hence
  $$\frac{1}{(x-2)(x-3)} = -\frac{1}{x-2} + \frac{1}{x-3}.$$
  So $$\int \frac{1}{(x-2)(x-3)}\,dx = -\ln|x-2| + \ln|x-3| = \ln\Bigl|\frac{x-3}{x-2}\Bigr|.$$ The third part therefore equals $$\tfrac{15}{2}\,\ln\Bigl|\dfrac{x-3}{x-2}\Bigr|.$$

5. Combine the results

$$I = x + \tfrac52\,\ln|x^2 - 5x + 6| + \tfrac{15}{2}\,\ln\Bigl|\dfrac{x-3}{x-2}\Bigr| + C,$$
where $$C$$ is the constant of integration.

Let
$$I = \displaystyle\int \frac{3x-2}{(x+1)^2\,(x+3)}\,dx.$$

Because the denominator contains the repeated linear factor $$x+1$$ and the distinct factor $$x+3$$, write a partial-fraction decomposition:

$$\frac{3x-2}{(x+1)^2(x+3)} = \frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+3}. $$

Multiply by the common denominator $$(x+1)^2(x+3)$$:

$$3x-2 = A(x+1)(x+3)+B(x+3)+C(x+1)^2.$$

Expand the right–hand side.

• $$A(x+1)(x+3)=A(x^2+4x+3)=A x^2+4A x+3A.$$
• $$B(x+3)=Bx+3B.$$
• $$C(x+1)^2=C(x^2+2x+1)=C x^2+2C x+C.$$

Adding,

$$ (A+C)x^2 + (4A+B+2C)x + (3A+3B+C). $$

Match coefficients with $$3x-2$$:

Substitute $$C=-A$$ in the remaining two equations.

From the x-coefficient:
$$4A+B+2(-A)=3\;\Longrightarrow\;2A+B=3\;\Longrightarrow\;B=3-2A.$$

From the constant term:

$$3A+3B+(-A)=-2\;\Longrightarrow\;-4A+3B=-2.$$

Insert $$B=3-2A$$:

$$-4A+3(3-2A)=-2\;\Longrightarrow\;-4A+9-6A=-2\;\Longrightarrow\;-10A=-11\;\Longrightarrow\;A=\dfrac{11}{4}.$$

Then

$$C=-A=-\dfrac{11}{4}, \qquad B=3-2A=3-\dfrac{11}{2}=-\dfrac{5}{2}.$$

So

$$\frac{3x-2}{(x+1)^2(x+3)}=\frac{11/4}{x+1}-\frac{5/2}{(x+1)^2}-\frac{11/4}{x+3}.$$

Integrate term by term:

$$I=\int \frac{11/4}{x+1}\,dx-\int \frac{5/2}{(x+1)^2}\,dx-\int \frac{11/4}{x+3}\,dx.$$

• $$\displaystyle\int \frac{11/4}{x+1}\,dx=\frac{11}{4}\ln|x+1|.$$
• $$\displaystyle\int \frac{5/2}{(x+1)^2}\,dx=\frac{5}{2}\int (x+1)^{-2}\,dx=\frac{5}{2}\bigl(-(x+1)^{-1}\bigr)=-\frac{5}{2(x+1)}.$$
• $$\displaystyle\int \frac{11/4}{x+3}\,dx=\frac{11}{4}\ln|x+3|.$$

Combine the signs carefully:

$$I=\frac{11}{4}\ln|x+1|+\frac{5}{2(x+1)}-\frac{11}{4}\ln|x+3|+C.$$

One may group the logarithms:

$$I=\frac{11}{4}\ln\Bigl|\frac{x+1}{x+3}\Bigr|+\frac{5}{2(x+1)}+C.$$

We have to evaluate the integral

$$I = \displaystyle\int \frac{x^2}{(x^2+1)(x^2+4)}\,dx.$$

Justification for a simplified partial fraction. In general, for two irreducible quadratic factors we would write $$\dfrac{x^2}{(x^2+1)(x^2+4)} = \dfrac{Ax+B}{x^2+1}+\dfrac{Cx+D}{x^2+4}.$$ However, the integrand depends only on $$x^2$$. Putting $$y = x^2$$ turns it into the rational function $$\dfrac{y}{(y+1)(y+4)}$$, whose partial-fraction decomposition has only constants. Hence we can write directly

$$\frac{x^2}{(x^2+1)(x^2+4)} = \frac{A}{x^2+1}+\frac{B}{x^2+4},$$

with $$A$$ and $$B$$ constants. (Equivalently, the odd terms in $$x$$ are absent because the integrand is even.)

Multiplying both sides by $$(x^2+1)(x^2+4)$$:

$$x^2 = A\,(x^2+4) + B\,(x^2+1).$$

Expanding and collecting:

$$x^2 = (A+B)\,x^2 + (4A+B).$$

Equating coefficients gives the linear system

$$A+B=1,\qquad 4A+B=0.$$

Subtracting the first equation from the second: $$3A=-1,$$ so $$A=-\tfrac13$$ and $$B = 1-A = \tfrac43.$$

Therefore

$$\frac{x^2}{(x^2+1)(x^2+4)} = -\frac{1}{3}\cdot\frac{1}{x^2+1} + \frac{4}{3}\cdot\frac{1}{x^2+4}.$$

Now integrate term by term using $$\int \dfrac{dx}{x^2+a^2}=\dfrac1a\arctan\!\left(\dfrac{x}{a}\right)+C$$:

• $$\displaystyle\int \frac{dx}{x^2+1}=\arctan x.$$
• $$\displaystyle\int \frac{dx}{x^2+4}=\tfrac12\arctan\!\left(\tfrac{x}{2}\right).$$

Combining:

$$I = -\frac13\,\arctan x + \frac43 \cdot \frac12\,\arctan\!\left(\frac{x}{2}\right) + C = -\frac13\,\arctan x + \frac23\,\arctan\!\left(\frac{x}{2}\right) + C,$$

where $$C$$ is the constant of integration.

Step 1 : Simplify the denominator

Use the Pythagorean identity $$\cos^2\phi = 1-\sin^2\phi$$:

$$5-\cos^2\phi-4\sin\phi = 5-(1-\sin^2\phi)-4\sin\phi$$

$$\qquad\qquad\qquad = 4+\sin^2\phi-4\sin\phi$$

$$\qquad\qquad\qquad = (\sin\phi-2)^2.$$

Hence

$$I = \int \dfrac{(3\sin\phi-2)\cos\phi}{(\sin\phi-2)^2}\,d\phi.$$

Step 2 : Choose a substitution

Let $$t = \sin\phi-2\;\;\Longrightarrow\;\;dt = \cos\phi\,d\phi.$$

Express the numerator in terms of t:

$$\sin\phi = t+2\;\;\Longrightarrow\;\;3\sin\phi-2 = 3(t+2)-2 = 3t+4.$$

Therefore

$$I = \int \dfrac{3t+4}{t^2}\,dt.$$

Step 3 : Integrate

Rewrite the integrand:

$$\dfrac{3t+4}{t^2} = \frac{3}{t}+\frac{4}{t^2}.$$

Integrate term by term:

$$\int \frac{3}{t}\,dt = 3\ln|t|,$$

$$\int \frac{4}{t^2}\,dt = 4\int t^{-2}\,dt = 4(-t^{-1}) = -\frac{4}{t}.$$

Thus

$$I = 3\ln|t| - \frac{4}{t} + C.$$

Step 4 : Back-substitute

Recall $$t = \sin\phi-2$$:

$$I = 3\ln|\sin\phi-2| - \frac{4}{\sin\phi-2} + C.$$

We decompose the integrand into partial fractions.

Assume

$$\frac{x^2+x+1}{(x+2)(x^2+1)}=\frac{A}{x+2}+\frac{Bx+C}{x^2+1}$$

Multiplying by $$(x+2)(x^2+1)$$ gives

$$x^2+x+1=A(x^2+1)+(Bx+C)(x+2).$$

Expand the right side:

$$A(x^2+1)=Ax^2+A,$$

$$(Bx+C)(x+2)=Bx^2+2Bx+Cx+2C=Bx^2+(2B+C)x+2C.$$

Add the two expressions:

$$(A+B)x^2+(2B+C)x+(A+2C).$$

Match coefficients with $$x^2+x+1$$:

Solve the system:

• From $$A+B=1\;\Rightarrow\;B=1-A.$$
• Insert in $$2B+C=1$$:
  $$2(1-A)+C=1\;\Rightarrow\;C=-1+2A.$$
• Insert $$A, C$$ in $$A+2C=1$$:
  $$A+2(-1+2A)=1\;\Rightarrow\;5A-2=1\;\Rightarrow\;A=\dfrac{3}{5}.$$
• Then $$B=1-A=\dfrac{2}{5},\qquad C=-1+2A=\dfrac{1}{5}.$$

Hence

$$\frac{x^2+x+1}{(x+2)(x^2+1)}=\frac{\tfrac{3}{5}}{x+2}+\frac{\tfrac{2}{5}x+\tfrac{1}{5}}{x^2+1}.$$

Integrate term-wise:

$$\int\frac{x^2+x+1\,dx}{(x+2)(x^2+1)}=\frac{3}{5}\int\frac{dx}{x+2}+\frac{2}{5}\int\frac{x\,dx}{x^2+1}+\frac{1}{5}\int\frac{dx}{x^2+1}.$$

Evaluate each integral:

• $$\int\frac{dx}{x+2}=\ln|x+2|.$$
• Use substitution $$u=x^2+1$$ in $$\int\frac{x}{x^2+1}dx$$:
  $$\int\frac{x\,dx}{x^2+1}=\frac12\ln(x^2+1).$$
• $$\int\frac{dx}{x^2+1}=\tan^{-1}x.$$

Combine the results:

$$\int\frac{x^2+x+1\,dx}{(x+2)(x^2+1)}=\frac{3}{5}\ln|x+2|+\frac{1}{5}\ln(x^2+1)+\frac{1}{5}\tan^{-1}x+C.$$

We have to evaluate the integral $$I = \int \frac{x}{(x+1)(x+2)}\,dx$$.

Step 1: Partial-fraction decomposition

Assume $$\frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$$ for constants $$A$$ and $$B$$.

Multiply by $$(x+1)(x+2)$$ to clear denominators:

$$x = A(x+2) + B(x+1).$$

Expand and collect like terms:

$$x = A x + 2A + B x + B = (A+B)x + (2A+B).$$

Equating coefficients,

• of $$x$$: $$A + B = 1,$$
• constant term: $$2A + B = 0.$$

From $$A + B = 1$$ we get $$B = 1 - A$$. Substitute in $$2A + B = 0$$:

$$2A + (1 - A) = 0 \;\Rightarrow\; A + 1 = 0 \;\Rightarrow\; A = -1.$$

Then $$B = 1 - (-1) = 2.$$

Step 2: Rewrite the integrand

$$\frac{x}{(x+1)(x+2)} = \frac{-1}{x+1} + \frac{2}{x+2}.$$

Step 3: Integrate term by term

$$I = \int \left( \frac{-1}{x+1} + \frac{2}{x+2} \right) dx = -\int \frac{dx}{x+1} + 2\int \frac{dx}{x+2}.$$

Using $$\int \frac{dx}{x+c} = \ln|x+c| + C$$,

$$I = -\ln|x+1| + 2\ln|x+2| + C.$$

Step 4: Optional simplification

The logarithms can be combined:

$$I = \ln\left| \frac{(x+2)^2}{x+1} \right| + C.$$

Hence the antiderivative is $$-\ln|x+1| + 2\ln|x+2| + C$$ (or any equivalent form).

Let $$I = \int \frac{1}{x^{2}-9}\,dx$$.

Factor the denominator:
$$x^{2}-9=(x-3)(x+3).$$

Resolve the rational function into partial fractions:

$$\frac{1}{x^{2}-9}=\frac{A}{x-3}+\frac{B}{x+3}.$$

Clear denominators by multiplying both sides by $$(x-3)(x+3):$$

$$1=A(x+3)+B(x-3).$$

Find the constants $$A$$ and $$B$$.

• Put $$x=3$$: $$1=A(3+3)=6A \;\Rightarrow\; A=\frac16.$$
• Put $$x=-3$$: $$1=B(-3-3)=-6B \;\Rightarrow\; B=-\frac16.$$

Thus

$$\frac{1}{x^{2}-9}=\frac{1/6}{x-3}-\frac{1/6}{x+3}.$$

Substitute back into the integral:

$$I=\int\left(\frac{1/6}{x-3}-\frac{1/6}{x+3}\right)dx.$$

Take the constant $$\frac16$$ outside each integral:

$$I=\frac16\int\frac{dx}{x-3}-\frac16\int\frac{dx}{x+3}.$$

Both integrals are of the form $$\int \frac{dx}{x-a}=\ln|x-a|+C$$, so

$$I=\frac16\ln|x-3|-\frac16\ln|x+3|+C.$$

Combine the logarithms:

$$I=\frac16\ln\left|\frac{x-3}{x+3}\right|+C.$$

Hence the required antiderivative is obtained.

We have to evaluate the integral

$$\displaystyle \int \dfrac{3x-1}{(x-1)(x-2)(x-3)}\,dx.$$

1. Decompose into partial fractions

Assume

$$\dfrac{3x-1}{(x-1)(x-2)(x-3)}=\dfrac{A}{x-1}+\dfrac{B}{x-2}+\dfrac{C}{x-3}.$$

Multiply both sides by $$(x-1)(x-2)(x-3)$$ to clear denominators:

$$3x-1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2).$$

Now determine the constants $A,B,C$ by convenient substitutions:

• Put $$x=1$$: $$3(1)-1=2=A(1-2)(1-3)=A(-1)(-2)=2A\;\Rightarrow\;A=1.$$
• Put $$x=2$$: $$3(2)-1=5=B(2-1)(2-3)=B(1)(-1)=-B\;\Rightarrow\;B=-5.$$
• Put $$x=3$$: $$3(3)-1=8=C(3-1)(3-2)=C(2)(1)=2C\;\Rightarrow\;C=4.$$

Hence

$$\dfrac{3x-1}{(x-1)(x-2)(x-3)}=\dfrac{1}{x-1}-\dfrac{5}{x-2}+\dfrac{4}{x-3}.$$

2. Integrate term-by-term

$$\int\dfrac{3x-1}{(x-1)(x-2)(x-3)}\,dx=\int\left(\dfrac{1}{x-1}-\dfrac{5}{x-2}+\dfrac{4}{x-3}\right)dx.$$

Each term is of the form $$\int \dfrac{1}{x-a}\,dx=\ln|x-a|+C$$, so

$$=\ln|x-1|-5\ln|x-2|+4\ln|x-3|+C.$$

3. Final result

$$\boxed{\displaystyle \int \dfrac{3x-1}{(x-1)(x-2)(x-3)}\,dx=\ln|x-1|-5\ln|x-2|+4\ln|x-3|+C.}$$

We have to evaluate $$I = \displaystyle\int \dfrac{x}{(x-1)(x-2)(x-3)}\,dx.$$

1. Decompose into partial fractions. Assume

$$\frac{x}{(x-1)(x-2)(x-3)} = \frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}.$$

Multiply both sides by $$(x-1)(x-2)(x-3)$$:

$$x = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2).$$

Determine $$A, B, C$$ by substituting the roots of the linear factors:

• Set $$x = 1$$:  $$1 = A(-1)(-2) = 2A \;\Rightarrow\; A = \dfrac12.$$
• Set $$x = 2$$:  $$2 = B(1)(-1) = -B \;\Rightarrow\; B = -2.$$
• Set $$x = 3$$:  $$3 = C(2)(1) = 2C \;\Rightarrow\; C = \dfrac32.$$

2. Write the integrand using the constants.

$$\frac{x}{(x-1)(x-2)(x-3)} = \frac{1/2}{x-1}-\frac{2}{x-2}+\frac{3/2}{x-3}.$$

3. Integrate term-wise. Using $$\int \dfrac{dx}{x-a}=\ln|x-a|$$:

$$I = \frac12\ln|x-1| - 2\ln|x-2| + \frac32\ln|x-3| + C,$$

where $$C$$ is the constant of integration.

We have to evaluate the integral

$$\int \frac{2x}{x^{2}+3x+2}\,dx.$$

Factor the denominator:

$$x^{2}+3x+2=(x+1)(x+2).$$

Decompose into partial fractions:

$$\frac{2x}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}.$$

Multiply both sides by $$(x+1)(x+2)$$:

$$2x=A(x+2)+B(x+1).$$

Substitute the roots of the linear factors:

• Set $$x=-1$$:  $$-2 = A(1) \;\Rightarrow\; A=-2.$$
• Set $$x=-2$$:  $$-4 = B(-1) \;\Rightarrow\; B=4.$$

Therefore

$$\frac{2x}{x^{2}+3x+2}=\frac{-2}{x+1}+\frac{4}{x+2}.$$

Now integrate term-wise using $$\int\dfrac{1}{x+k}\,dx=\ln|x+k|$$:

$$\int\frac{2x}{x^{2}+3x+2}\,dx=-2\ln|x+1|+4\ln|x+2|+C.$$

We have to integrate the rational function $$\dfrac{1 - x^2}{x(1 - 2x)}.$$

Because the degree of the numerator equals the degree of the denominator, start with polynomial long-division to obtain a proper fraction.

Divide $$1 - x^{2}$$ by $$x(1-2x)=-2x^{2}+x$$. The leading terms are $$-x^{2}$$ and $$-2x^{2}$$, so the first term of the quotient is $$\dfrac{-x^{2}}{-2x^{2}}=\dfrac12.$$

Multiply the divisor by $$\dfrac12$$ and subtract:

$$\bigl(1 - x^{2}\bigr)-\dfrac12\bigl(-2x^{2}+x\bigr)=1-x^{2}+x^{2}-\dfrac x2=1-\dfrac x2.$$

Thus

$$\dfrac{1 - x^{2}}{x(1-2x)}=\dfrac12+\dfrac{1-\dfrac x2}{x(1-2x)}.$$

The remaining fraction is proper. Decompose into partial fractions:

$$\dfrac{1-\dfrac x2}{x(1-2x)}=\dfrac{A}{x}+\dfrac{B}{1-2x}.$$

Clearing the denominators:

$$A(1-2x)+Bx=1-\dfrac x2.$$

Equate coefficients of like powers of $$x$$:

• Constant term: $$A=1.$$
• Coefficient of $$x$$: $$-2A+B=-\dfrac12 \;\Longrightarrow\; B = -\dfrac12+2 = \dfrac32.$$

Hence

$$\dfrac{1-\dfrac x2}{x(1-2x)}=\dfrac1x+\dfrac{3/2}{1-2x}.$$

Substituting back, the integrand becomes

$$\dfrac{1 - x^{2}}{x(1 - 2x)}=\dfrac12+\dfrac1x+\dfrac{3/2}{1-2x}.$$

Now integrate term by term:

• $$\displaystyle \int\dfrac12\,dx=\dfrac x2.$$
• $$\displaystyle \int\dfrac1x\,dx=\ln|x|.$$
• For $$\displaystyle\int\dfrac{3/2}{1-2x}\,dx$$, use $$\int\dfrac{1}{a+bx}\,dx=\dfrac1b\ln|a+bx|$$ with $$a=1$$, $$b=-2$$: $$\int\dfrac{3/2}{1-2x}\,dx=\dfrac32\cdot\left(-\dfrac12\right)\ln|1-2x|=-\dfrac34\ln|1-2x|.$$

Adding the three results:

$$\int\dfrac{1 - x^{2}}{x(1 - 2x)}\,dx=\dfrac x2+\ln|x|-\dfrac34\ln|1-2x|+C.$$

Integral to be evaluated
Find $$\displaystyle \int \dfrac{x\,dx}{(x^2+1)(x-1)}.$$

Step 1: Decompose into partial fractions

Because $$x^2+1$$ is an irreducible quadratic and $$x-1$$ is linear, write

$$\dfrac{x}{(x^2+1)(x-1)}=\dfrac{Ax+B}{x^2+1}+\dfrac{C}{x-1}.$$

Multiply both sides by $$(x^2+1)(x-1)$$ to clear denominators:

$$x=(Ax+B)(x-1)+C(x^2+1).$$

Expand the right–hand side

$$(Ax+B)(x-1)=A x^2+(B-A)x-B,$$ so

$$(Ax+B)(x-1)+C(x^2+1)=(A+C)x^2+(B-A)x-B+C.$$

Equate coefficients with the left side $$x=0\,x^2+1\,x+0$$:

Insert $$C=-A$$ into the constant-term equation:

$$-B-A=0\;\;\Rightarrow\;\;B=-A.$$

Now use $$B-A=1$$:

$$(-A)-A=1\;\;\Rightarrow\;\;-2A=1\;\;\Rightarrow\;\;A=-\dfrac12.$$

Hence $$B=-A=\dfrac12$$ and $$C=-A=\dfrac12.$$

Step 2: Rewrite the integrand

$$\dfrac{x}{(x^2+1)(x-1)}=\dfrac{-\tfrac12 x+\tfrac12}{x^2+1}+\dfrac{\tfrac12}{x-1}.$$

Separate the terms inside the integral:

$$\int \dfrac{x\,dx}{(x^2+1)(x-1)}=\int\!\left(\frac{-\tfrac12 x}{x^2+1}+\frac{\tfrac12}{x^2+1}+\frac{\tfrac12}{x-1}\right)dx.$$

Step 3: Integrate each part

1. For $$\displaystyle \int \frac{-\tfrac12 x}{x^2+1}\,dx$$ place $$u=x^2+1\;\Rightarrow\;du=2x\,dx.$$ Then $$x\,dx=\dfrac{du}{2},$$ so $$\int \frac{-\tfrac12 x}{x^2+1}\,dx=-\tfrac12\int\frac{x\,dx}{x^2+1}=-\tfrac12\int\frac{du/2}{u}=-\tfrac14\ln|u|=-\tfrac14\ln(x^2+1).$$
2. For $$\displaystyle \int \frac{\tfrac12}{x^2+1}\,dx$$ use the standard form $$\int \frac{dx}{x^2+1}=\tan^{-1}x,$$ giving $$\frac12\tan^{-1}x.$$
3. For $$\displaystyle \int \frac{\tfrac12}{x-1}\,dx$$ we obtain $$\tfrac12\ln|x-1|.$$

Step 4: Collect the results

$$\int \dfrac{x\,dx}{(x^2+1)(x-1)}=-\tfrac14\ln(x^2+1)+\tfrac12\tan^{-1}x+\tfrac12\ln|x-1|+C,$$

where $$C$$ is the constant of integration.

We have to evaluate the integral

$$I = \int \frac{x}{(x-1)^2\,(x+2)}\,dx$$

1. Express the integrand as partial fractions

Because the denominator already factors into linear terms, we assume

$$\frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}.$$

Multiply both sides by the common denominator $$(x-1)^2(x+2)$$ to clear fractions:

$$x = A(x-1)(x+2) + B(x+2) + C(x-1)^2.$$

Expand each product:

• $$(x-1)(x+2) = x^2 + x - 2\;\Rightarrow\;A(x-1)(x+2)=A x^2 + A x - 2A.$$
• $$B(x+2)=B x + 2B.$$
• $$(x-1)^2 = x^2 - 2x + 1\;\Rightarrow\;C(x-1)^2 = C x^2 - 2C x + C.$$

Add the like terms:

$$x = (A + C) x^2 + (A + B - 2C) x + (-2A + 2B + C).$$

Equate coefficients with the left-hand side $$0\,x^2 + 1\,x + 0$$:

Solve this system:

• From $$A + C = 0$$, we get $$C = -A.$$
• Insert $$C=-A$$ into the other two equations.

Second equation:

$$A + B - 2(-A) = A + B + 2A = 3A + B = 1\;\Rightarrow\;B = 1 - 3A.$$

Third equation:

$$-2A + 2B + (-A) = -3A + 2B = 0\;\Rightarrow\;2B = 3A\;\Rightarrow\;B = \frac{3A}{2}.$$

Equate the two expressions for $$B$$:

$$1 - 3A = \frac{3A}{2}\;\Longrightarrow\;2 - 6A = 3A\;\Longrightarrow\;9A = 2\;\Longrightarrow\;A = \frac{2}{9}.$$

Then

$$B = \frac{3A}{2} = \frac{3}{2}\cdot\frac{2}{9} = \frac{1}{3},\qquad C = -A = -\frac{2}{9}.$$

Hence

$$\frac{x}{(x-1)^2(x+2)} = \frac{\tfrac{2}{9}}{x-1} + \frac{\tfrac{1}{3}}{(x-1)^2} - \frac{\tfrac{2}{9}}{x+2}.$$

2. Integrate term by term

$$I = \int \Bigg[ \frac{2}{9}\,\frac{1}{x-1} + \frac{1}{3}\,\frac{1}{(x-1)^2} - \frac{2}{9}\,\frac{1}{x+2} \Bigg] dx.$$

• $$\int \frac{1}{x-1}\,dx = \ln|x-1|.$$
• $$\int \frac{1}{(x-1)^2}\,dx = -\frac{1}{x-1}.$$ (Differentiate to verify.)
• $$\int \frac{1}{x+2}\,dx = \ln|x+2|.$$

Applying the constants:

$$I = \frac{2}{9}\ln|x-1| + \frac{1}{3}\Bigl(-\frac{1}{x-1}\Bigr) - \frac{2}{9}\ln|x+2| + C.$$

3. Final result

$$I = \boxed{\frac{2}{9}\ln|x-1|\; - \;\frac{1}{3(x-1)}\; - \;\frac{2}{9}\ln|x+2| + C}.$$

We have to evaluate the integral $$\displaystyle\int \frac{3x+5}{x^{3}-x^{2}-x+1}\,dx$$.

1. Factorise the denominator

Group the terms:
$$x^{3}-x^{2}-x+1 = x^{2}(x-1)-1(x-1)= (x^{2}-1)(x-1).$$
Since $$x^{2}-1=(x-1)(x+1)$$, the complete factorisation is
$$x^{3}-x^{2}-x+1=(x-1)^{2}(x+1).$$

2. Set up the partial-fraction form

Because the denominator has a repeated linear factor \((x-1)^{2}\) and a simple linear factor \((x+1)\), write

$$\frac{3x+5}{(x-1)^{2}(x+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+1}.$$

3. Determine the constants \(A,\,B,\,C\)

Multiply by the denominator to clear fractions:

$$3x+5 = A(x-1)(x+1)+B(x+1)+C(x-1)^{2}.$$

Expand every term.

• $$A(x-1)(x+1)=A(x^{2}-1)=A x^{2}-A.$$
• $$B(x+1)=Bx+B.$$
• $$C(x-1)^{2}=C(x^{2}-2x+1)=C x^{2}-2C x+C.$$

Add the like terms:

$$3x+5=(A+C)x^{2}+(B-2C)x+(-A+B+C).$$

Now equate coefficients of the powers of \(x\):

Solve the linear system:

From $$A+C=0\;\Rightarrow\;C=-A.$$

Substitute in the second equation:
$$B-2(-A)=B+2A=3\;\Rightarrow\;B=3-2A.$$

Substitute \(B\) and \(C\) into the third equation:

$$-A+(3-2A)+(-A)=5 \;\Longrightarrow\; 3-4A=5\;\Longrightarrow\; -4A=2\;\Longrightarrow\; A=-\tfrac12.$$

Hence
$$C=-A=\tfrac12,\qquad B=3-2A=3-2(-\tfrac12)=4.$$

4. Write the decomposition

$$\frac{3x+5}{x^{3}-x^{2}-x+1}=\frac{-\tfrac12}{x-1}+\frac{4}{(x-1)^{2}}+\frac{\tfrac12}{x+1}.$$

5. Integrate term by term

\(\displaystyle\int \! \frac{-\tfrac12}{x-1}\,dx = -\tfrac12\,\ln|x-1|.\)

\(\displaystyle\int \! \frac{4}{(x-1)^{2}}\,dx = 4\int (x-1)^{-2}\,dx =4\bigl[-(x-1)^{-1}\bigr]= -\frac{4}{x-1}.\)

\(\displaystyle\int \! \frac{\tfrac12}{x+1}\,dx = \tfrac12\,\ln|x+1|.\)

6. Combine the results

Therefore

$$\int \frac{3x+5}{x^{3}-x^{2}-x+1}\,dx = -\tfrac12\,\ln|x-1| - \frac{4}{x-1} + \tfrac12\,\ln|x+1| + C.$$

The two logarithms can be grouped:

$$\boxed{\displaystyle \int \frac{3x+5}{x^{3}-x^{2}-x+1}\,dx = \tfrac12\,\ln\left|\frac{x+1}{x-1}\right| - \frac{4}{x-1} + C}. $$

Let us integrate

$$I = \int \dfrac{2x - 3}{(x^2-1)(2x+3)}\,dx.$$

First factorise the quadratic in the denominator:

$$x^2 - 1=(x-1)(x+1).$$

Hence

$$\dfrac{2x-3}{(x^2-1)(2x+3)}=\dfrac{2x-3}{(x-1)(x+1)(2x+3)}.$$

Because the three factors are distinct and linear, decompose into partial fractions:

$$\frac{2x-3}{(x-1)(x+1)(2x+3)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{2x+3}.$$

Multiply by the common denominator \((x-1)(x+1)(2x+3)\) to obtain the identity

$$2x-3=A(x+1)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+1).$$

Determine the unknown constants by suitable choices of \(x\):

• Put $$x=1$$:

  $$2(1)-3=-1=10A \;\Rightarrow\; A=-\dfrac1{10}.$$

• Put $$x=-1$$:

  $$2(-1)-3=-5=-2B \;\Rightarrow\; B=\dfrac52.$$

• Put $$x=-\dfrac32$$ (so that \(2x+3=0\)):

  $$2\left(-\dfrac32\right)-3=-6=\frac54\,C \;\Rightarrow\; C=-\dfrac{24}{5}. $$

The partial-fraction form therefore is

$$\frac{2x-3}{(x^2-1)(2x+3)}=-\frac1{10}\,\frac1{x-1}+\frac52\,\frac1{x+1}-\frac{24}{5}\,\frac1{2x+3}.$$

Now integrate term by term:

\[ \begin{aligned} I &=\int\left(-\frac1{10}\,\frac1{x-1}\right)dx+\int\left(\frac52\,\frac1{x+1}\right)dx+\int\left(-\frac{24}{5}\,\frac1{2x+3}\right)dx\\[4pt] &=-\frac1{10}\,\ln|x-1|+\frac52\,\ln|x+1|-\frac{24}{5}\cdot\frac12\,\ln|2x+3|+C\\[4pt] &=-\frac1{10}\ln|x-1|+\frac52\ln|x+1|-\frac{12}{5}\ln|2x+3|+C. \end{aligned} \]

Thus the required integral is

$$I=-\dfrac1{10}\,\ln|x-1|+\dfrac52\,\ln|x+1|-\dfrac{12}{5}\,\ln|2x+3|+C.$$

We have to evaluate the integral

$$I = \displaystyle\int\frac{5x}{(x+1)(x^{2}-4)}\,dx = \int\frac{5x}{(x+1)(x-2)(x+2)}\,dx.$$

1. Decompose into partial fractions

Assume

$$\frac{5x}{(x+1)(x-2)(x+2)} = \frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{x+2}.$$

Multiply both sides by $$(x+1)(x-2)(x+2)$$ to clear denominators:

$$5x = A(x-2)(x+2) + B(x+1)(x+2) + C(x+1)(x-2).$$

Compute each product:

• $$(x-2)(x+2) = x^{2}-4.$$
• $$(x+1)(x+2) = x^{2}+3x+2.$$
• $$(x+1)(x-2) = x^{2}-x-2.$$

Substitute back:

$$5x = A(x^{2}-4) + B(x^{2}+3x+2) + C(x^{2}-x-2).$$

Group like powers of $$x$$:

$$5x = (A+B+C)x^{2} + (3B - C)x + (-4A + 2B - 2C).$$

Match coefficients with $$0x^{2} + 5x + 0$$ to obtain the system

1. $$A + B + C = 0$$
2. $$3B - C = 5$$
3. $$-4A + 2B - 2C = 0$$

From (1) $$C = -A - B.$$ Substitute this value of $$C$$ in (2) and (3):

(2) $$3B - ( -A - B ) = 5 \;\Rightarrow\; A + 4B = 5.$$

(3) $$-4A + 2B - 2(-A - B) = 0 \;\Rightarrow\; -2A + 4B = 0$$ giving $$A = 2B.$$

Insert $$A = 2B$$ into $$A + 4B = 5$$:

$$2B + 4B = 5 \;\Rightarrow\; 6B = 5 \;\Rightarrow\; B = \frac56.$$

Hence

$$A = 2B = 2\left(\frac56\right) = \frac53,\qquad C = -A - B = -\frac53 - \frac56 = -\frac52.$$

2. Write the integrand with these constants

$$\frac{5x}{(x+1)(x^{2}-4)} = \frac{\tfrac53}{x+1} + \frac{\tfrac56}{x-2} - \frac{\tfrac52}{x+2}.$$

3. Integrate term by term

$$\begin{aligned} I &= \int \frac{\tfrac53}{x+1}\,dx + \int \frac{\tfrac56}{x-2}\,dx - \int \frac{\tfrac52}{x+2}\,dx \\ &= \frac53\ln|x+1| + \frac56\ln|x-2| - \frac52\ln|x+2| + C. \end{aligned}$$

(A common factor $$5$$ may be taken outside if desired.)

4. Final result

$$\boxed{\displaystyle\int\frac{5x}{(x+1)(x^{2}-4)}\,dx = \frac53\ln|x+1| + \frac56\ln|x-2| - \frac52\ln|x+2| + C}$$

Let $$I = \int \dfrac{x^3 + x + 1}{x^2 - 1}\,dx$$.

1. Divide the polynomials

Divide the numerator $$x^3 + x + 1$$ by the denominator $$x^2 - 1$$:

$$x^3 + x + 1 = x\,(x^2 - 1) + (2x + 1).$$

Therefore

$$\dfrac{x^3 + x + 1}{x^2 - 1} = x + \dfrac{2x + 1}{x^2 - 1}.$$

2. Resolve the proper fraction into partial fractions

Since $$x^2 - 1 = (x - 1)(x + 1),$$ write

$$\dfrac{2x + 1}{x^2 - 1} = \dfrac{A}{x - 1} + \dfrac{B}{x + 1}.$$

Combine the right-hand side:

$$\dfrac{A(x + 1) + B(x - 1)}{(x - 1)(x + 1)} = \dfrac{(A + B)x + (A - B)}{x^2 - 1}.$$

Equating numerators,

• coefficient of $$x$$: $$A + B = 2,$$
• constant term: $$A - B = 1.$$

Solving,

$$A = \tfrac{3}{2}, \qquad B = \tfrac{1}{2}.$$

Thus

$$\dfrac{x^3 + x + 1}{x^2 - 1} = x + \dfrac{\tfrac{3}{2}}{x - 1} + \dfrac{\tfrac{1}{2}}{x + 1}.$$

3. Integrate term by term

$$\begin{aligned} I &= \int x\,dx + \frac{3}{2}\int \frac{dx}{x - 1} + \frac{1}{2}\int \frac{dx}{x + 1}\\[4pt] &= \frac{x^2}{2} + \frac{3}{2}\ln|x - 1| + \frac{1}{2}\ln|x + 1| + C. \end{aligned}$$

4. Result

Hence $$\displaystyle \int \dfrac{x^3 + x + 1}{x^2 - 1}\,dx = \dfrac{x^2}{2} + \dfrac{3}{2}\ln|x - 1| + \dfrac{1}{2}\ln|x + 1| + C.$$

The required integral is

$$I = \int \dfrac{2}{(1-x)(1+x^2)}\,dx$$

1. Express the integrand in partial fractions

Assume

$$\dfrac{2}{(1-x)(1+x^2)} = \dfrac{A}{1-x} + \dfrac{Bx + C}{1+x^2}$$

Multiply both sides by $$(1-x)(1+x^2)$$:

$$2 = A(1+x^2) + (Bx + C)(1-x)$$

Expand the right–hand side:

$$A(1+x^2) + (Bx + C) - (Bx + C)x$$

$$= A + Ax^2 + Bx + C - Bx^2 - Cx$$

Group like terms:

• Constant term: $$A + C$$
• Coefficient of $$x$$: $$B - C$$
• Coefficient of $$x^2$$: $$A - B$$

Since the left–hand side is the constant $$2$$, compare coefficients:

$$A - B = 0 \;\Rightarrow\; A = B$$

$$B - C = 0 \;\Rightarrow\; C = B$$

$$A + C = 2$$

With $$A = B$$ and $$C = B$$, the last equation gives $$2 = A + A = 2A \Rightarrow A = 1$$. Thus

$$A = B = C = 1$$

Hence

$$\dfrac{2}{(1-x)(1+x^2)} = \dfrac{1}{1-x} + \dfrac{x+1}{1+x^2}$$

2. Integrate term-by-term

$$I = \int \dfrac{1}{1-x}\,dx + \int \dfrac{x}{1+x^2}\,dx + \int \dfrac{1}{1+x^2}\,dx$$

1. First integral: let $$u = 1 - x$$, so $$du = -dx$$.

$$\int \dfrac{1}{1-x}\,dx = -\ln|1-x|$$

2. Second integral: let $$u = 1 + x^2$$, so $$du = 2x\,dx$$.

$$\int \dfrac{x}{1+x^2}\,dx = \frac12 \ln(1+x^2)$$

3. Third integral (standard form):

$$\int \dfrac{1}{1+x^2}\,dx = \arctan x$$

3. Combine the results

$$I = -\ln|1-x| + \frac12\,\ln(1+x^2) + \arctan x + C$$

where $$C$$ is the constant of integration.

Given integral

We have to evaluate $$I = \displaystyle\int \frac{3x-1}{(x+2)^2}\,dx.$$

Step 1: Rewrite the numerator

Write $$3x-1$$ in terms of $$(x+2):$$

$$\begin{aligned} 3x-1 &= 3\bigl[(x+2)-2\bigr]-1 \\[-2pt] &= 3(x+2)-6-1 \\[-2pt] &= 3(x+2)-7. \end{aligned}$$

Step 2: Split the fraction

Substituting the expression just found,

$$\frac{3x-1}{(x+2)^2}=\frac{3(x+2)-7}{(x+2)^2}=\frac{3}{x+2}-\frac{7}{(x+2)^2}.$$

Step 3: Integrate term-wise

$$\begin{aligned} I &= \int\left(\frac{3}{x+2}-\frac{7}{(x+2)^2}\right)dx \\[4pt] &= 3\int\frac{dx}{x+2}-7\int\frac{dx}{(x+2)^2}. \end{aligned}$$

• First integral:
  $$3\int\frac{dx}{x+2}=3\ln|x+2|.$$
• Second integral:
  Put $$t=x+2\;(dt=dx).$$ Then
  $$-7\int\frac{dx}{(x+2)^2}=-7\int t^{-2}dt=-7\left(-t^{-1}\right)=\frac{7}{t}=\frac{7}{x+2}.$$

Step 4: Combine the results

$$I = 3\ln|x+2| + \frac{7}{x+2} + C,$$

where $$C$$ is the constant of integration.

Required integral

$$I = \int \dfrac{dx}{x^{4}-1}$$

1. Factor the denominator

$$x^{4}-1 = (x^{2}+1)(x^{2}-1) = (x^{2}+1)(x+1)(x-1).$$

2. Write the partial-fraction decomposition. Because $$x-1$$ and $$x+1$$ are linear and $$x^{2}+1$$ is irreducible over $$\mathbb{R}$$,

$$\dfrac{1}{x^{4}-1}=\dfrac{A}{x-1}+\dfrac{B}{x+1}+\dfrac{Cx+D}{x^{2}+1}.$$

3. Determine the constants. Multiply both sides by $$x^{4}-1$$:

$$1=A(x+1)(x^{2}+1)+B(x-1)(x^{2}+1)+(Cx+D)(x^{2}-1).$$

Substitute the real roots of the linear factors:

• Set $$x=1$$:  $$1 = A(2)(2) = 4A \;\Rightarrow\; A = \tfrac14.$$
• Set $$x=-1$$:  $$1 = B(-2)(2) = -4B \;\Rightarrow\; B = -\tfrac14.$$

To find $$C, D$$, expand:

$$1=(A+B+C)x^{3}+(A-B+D)x^{2}+(A+B-C)x+(A-B-D).$$

Equating coefficients with $$1 = 0\,x^3 + 0\,x^2 + 0\,x + 1$$:

• $$x^{3}$$: $$A+B+C=0 \;\Rightarrow\; C = 0$$ (since $$A+B=0$$).
• Constant: $$A-B-D=1 \;\Rightarrow\; D = A-B-1 = \tfrac14 + \tfrac14 - 1 = -\tfrac12.$$

4. Write the decomposition with the found constants.

$$\dfrac{1}{x^{4}-1}=\dfrac{1}{4}\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}\right)-\dfrac12\cdot\dfrac{1}{x^{2}+1}.$$

5. Integrate term by term.

$$I = \dfrac14\int\!\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}\right)dx - \dfrac12\int\dfrac{dx}{x^{2}+1} = \dfrac14\bigl[\ln|x-1|-\ln|x+1|\bigr]-\dfrac12\tan^{-1}x+C.$$

6. Combine the logarithms.

$$I = \dfrac14\ln\left|\dfrac{x-1}{x+1}\right|-\dfrac12\tan^{-1}x+C.$$

We have to evaluate the integral

$$I = \int \dfrac{1}{x\,(x^{n}+1)}\,dx.$$

Step 1 – Multiply numerator and denominator by $$x^{n-1}$$.

$$I = \int \dfrac{x^{n-1}}{x^{n}(x^{n}+1)}\,dx.$$

Step 2 – Put $$x^{n}=t$$.

• Then $$dt = n\,x^{n-1}\,dx\;\;\Rightarrow\;\;x^{n-1}\,dx = \dfrac{dt}{n}$$.

Substituting,

$$I = \int \dfrac{\dfrac{dt}{n}}{t\,(t+1)} = \dfrac1n\int \dfrac{1}{t(t+1)}\,dt.$$

Step 3 – Resolve into partial fractions.

$$\dfrac{1}{t(t+1)} = \dfrac{1}{t} - \dfrac{1}{t+1}.$$ Therefore,

$$I = \dfrac1n \int \left( \dfrac{1}{t} - \dfrac{1}{t+1} \right) dt.$$

Step 4 – Integrate.

$$I = \dfrac1n \left( \ln|t| - \ln|t+1| \right) + C.$$

Step 5 – Re-substitute $$t = x^{n}$$.

$$I = \dfrac1n \ln\left| \dfrac{x^{n}}{x^{n}+1} \right| + C.$$

Hence, the required integral is

$$\boxed{\displaystyle \int \dfrac{dx}{x\,(x^{n}+1)} = \dfrac1n\,\ln\left| \dfrac{x^{n}}{x^{n}+1} \right| + C}.$$

We have to evaluate the integral

$$I = \displaystyle\int \dfrac{\cos x}{(1-\sin x)(2-\sin x)}\,dx.$$

Step 1   Substitution

Put $$\sin x = t\;,$$ so that $$dt = \cos x\,dx \;\Rightarrow\; \cos x\,dx = dt.$$
The integral becomes

$$I = \int \dfrac{dt}{(1-t)(2-t)}.$$

Step 2   Partial-fraction decomposition

Write

$$\frac{1}{(1-t)(2-t)} = \frac{A}{1-t} + \frac{B}{2-t}.$$

Multiplying by $$(1-t)(2-t)$$ gives

$$1 = A(2-t) + B(1-t).$$

Comparing coefficients of powers of $$t$$:

• Coefficient of $$t$$: $$-A - B = 0 \;\Rightarrow\; A + B = 0.$$
• Constant term: $$2A + B = 1.$$

Solving: from $$B = -A$$ substitute into $$2A + B = 1$$ to get $$2A - A = 1 \;\Rightarrow\; A = 1$$, hence $$B = -1.$$

Thus

$$\frac{1}{(1-t)(2-t)} = \frac{1}{1-t} - \frac{1}{2-t}.$$

Step 3   Integrate term-by-term

Therefore

$$I = \int \left[\frac{1}{1-t} - \frac{1}{2-t}\right] dt$$

$$\phantom{I} = \int \frac{1}{1-t}\,dt \;\; - \;\; \int \frac{1}{2-t}\,dt.$$

We use the standard result $$\int \frac{1}{a-t}\,dt = -\ln|a-t| + C.$$ Hence

$$I = \Bigl[-\ln|1-t|\Bigr] - \Bigl[-\ln|2-t|\Bigr] + C$$

$$\phantom{I} = -\ln|1-t| + \ln|2-t| + C$$

$$\phantom{I} = \ln\left|\frac{2-t}{1-t}\right| + C.$$

Step 4   Back-substitute $$t = \sin x$$

Finally,

$$\boxed{\displaystyle \int \frac{\cos x}{(1-\sin x)(2-\sin x)}\,dx = \ln\left|\frac{2-\sin x}{1-\sin x}\right| + C.}$$

We have to evaluate

$$I = \int \dfrac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}\,dx.$$

1. Expand numerator and denominator.

$$ (x^2+1)(x^2+2)=x^4+3x^2+2, \qquad (x^2+3)(x^2+4)=x^4+7x^2+12.$$

2. Perform long division. Both quartics have the same degree, so

$$\dfrac{x^4+3x^2+2}{x^4+7x^2+12}=1+\dfrac{(x^4+3x^2+2)-(x^4+7x^2+12)}{x^4+7x^2+12}=1+\dfrac{-4x^2-10}{(x^2+3)(x^2+4)}.$$

Hence

$$I = \int 1\,dx + \int \dfrac{-4x^2-10}{(x^2+3)(x^2+4)}\,dx.$$

3. Partial-fraction decomposition of the proper part. Since the integrand depends only on $$x^2$$, write

$$\dfrac{-4x^2-10}{(x^2+3)(x^2+4)} = \dfrac{A}{x^2+3}+\dfrac{B}{x^2+4}.$$

Multiply by the denominator:

$$-4x^2-10 = A(x^2+4)+B(x^2+3)=(A+B)x^2+(4A+3B).$$

Equating coefficients gives the system

$$A+B=-4,\qquad 4A+3B=-10.$$

From the first, $$B=-4-A$$. Substituting: $$4A+3(-4-A)= A-12 = -10$$, so $$A=2$$ and $$B=-6.$$

Therefore

$$\dfrac{-4x^2-10}{(x^2+3)(x^2+4)}=\frac{2}{x^2+3}-\frac{6}{x^2+4}.$$

4. Rewrite the integrand.

$$\dfrac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}=1+\frac{2}{x^2+3}-\frac{6}{x^2+4}.$$

5. Integrate term by term using $$\int \dfrac{dx}{x^2+a^2}=\dfrac{1}{a}\tan^{-1}\!\left(\dfrac{x}{a}\right)$$:

• $$\int 1\,dx = x.$$
• $$\int \frac{2}{x^2+3}\,dx = \frac{2}{\sqrt{3}}\tan^{-1}\!\left(\dfrac{x}{\sqrt{3}}\right).$$
• $$\int \frac{-6}{x^2+4}\,dx = -3\tan^{-1}\!\left(\dfrac{x}{2}\right).$$

6. Assemble the result.

$$I=x+\frac{2}{\sqrt{3}}\tan^{-1}\!\left(\dfrac{x}{\sqrt{3}}\right)-3\tan^{-1}\!\left(\dfrac{x}{2}\right)+C.$$

We have to evaluate the integral

$$I=\int \dfrac{2x}{(x^{2}+1)(x^{2}+3)}\,dx.$$

Step 1: Substitution. Put $$y=x^{2}$$ in the integrand; then $$dy = 2x\,dx$$, so $$2x\,dx = dy$$ and

$$I = \int \dfrac{dy}{(y+1)(y+3)}.$$

Step 2: Partial-fraction decomposition.

$$\dfrac{1}{(y+1)(y+3)} = \dfrac{A}{y+1}+\dfrac{B}{y+3}.$$

Multiplying out: $$1 = A(y+3)+B(y+1).$$ Substituting roots:

• Set $$y=-1$$: $$1 = 2A \Rightarrow A = \tfrac12.$$
• Set $$y=-3$$: $$1 = -2B \Rightarrow B = -\tfrac12.$$

Hence

$$I = \tfrac12 \int \dfrac{dy}{y+1} - \tfrac12\int \dfrac{dy}{y+3} = \tfrac12\ln|y+1| - \tfrac12\ln|y+3| + C.$$

Step 3: Back-substitute $$y = x^{2}$$:

$$I = \tfrac12\ln(x^{2}+1) - \tfrac12\ln(x^{2}+3) + C = \tfrac12\ln\!\left(\dfrac{x^{2}+1}{x^{2}+3}\right) + C.$$

We have to evaluate the integral

$$\displaystyle \int \frac{1}{x\,(x^{4}-1)}\,dx$$

First, factorise the denominator:

$$x^{4}-1=(x^{2}-1)(x^{2}+1)=(x-1)(x+1)(x^{2}+1)$$

Hence

$$\frac{1}{x(x^{4}-1)}=\frac{1}{x(x-1)(x+1)(x^{2}+1)}$$

Express the rational function in partial fractions.

Take

$$\frac{1}{x(x-1)(x+1)(x^{2}+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}+\frac{Dx+E}{x^{2}+1}$$

Multiply both sides by the common denominator to obtain the identity

$$1=A(x-1)(x+1)(x^{2}+1)+B\,x(x+1)(x^{2}+1)+C\,x(x-1)(x^{2}+1)+(Dx+E)\,x(x-1)(x+1)$$

Put convenient values of $$x$$ to find some of the constants.

• For $$x=0$$: $$1=-A\;\Rightarrow\;A=-1$$.
• For $$x=1$$: $$1=4B\;\Rightarrow\;B=\tfrac14$$.
• For $$x=-1$$: $$1=4C\;\Rightarrow\;C=\tfrac14$$.

We still need $$D,E$$. Substitute $$A,B,C$$ obtained above and simplify.

With $$A=-1$$ we have

$$\begin{aligned} T_1&=-1\,(x-1)(x+1)(x^{2}+1)=-(x^{2}-1)(x^{2}+1)=-(x^{4}-1)=-x^{4}+1,\\ T_2&=\tfrac14\,x(x+1)(x^{2}+1),\\ T_3&=\tfrac14\,x(x-1)(x^{2}+1). \end{aligned}$$

Add $$T_2$$ and $$T_3$$ first:

$$T_2+T_3=\tfrac14\,x(x^{2}+1)[(x+1)+(x-1)]=\tfrac12\,x^{2}(x^{2}+1)=\tfrac12x^{4}+\tfrac12x^{2}.$$

Thus

$$T_1+T_2+T_3=-x^{4}+1+\tfrac12x^{4}+\tfrac12x^{2}= -\tfrac12x^{4}+\tfrac12x^{2}+1.$$

We require $$T_1+T_2+T_3+T_4\equiv1$$, where

$$T_4=(Dx+E)\,x(x-1)(x+1)=(Dx+E)x(x^{2}-1)=(x^{2}-1)(Dx^{2}+Ex).$$

So $$T_4$$ must supply

$$T_4=\tfrac12x^{4}-\tfrac12x^{2}=\tfrac12x^{2}(x^{2}-1).$$

Comparing with $$T_4=(x^{2}-1)(Dx^{2}+Ex)$$ gives

$$Dx^{2}+Ex=\tfrac12x^{2} \Longrightarrow D=\tfrac12,\;E=0.$$

Therefore

$$\frac{1}{x(x^{4}-1)}=-\frac1x+\frac{1}{4(x-1)}+\frac{1}{4(x+1)}+\frac{\tfrac12x}{x^{2}+1}.$$

Integrate term by term:

$$\int-\frac1x\,dx=-\ln|x|+C_1$$

$$\int\frac{1}{4(x-1)}\,dx=\tfrac14\ln|x-1|+C_2$$

$$\int\frac{1}{4(x+1)}\,dx=\tfrac14\ln|x+1|+C_3$$

For the last term substitute $$u=x^{2}+1\;(du=2x\,dx)$$:

$$\int\frac{\tfrac12x}{x^{2}+1}\,dx=\tfrac12\int\frac{x\,dx}{x^{2}+1}=\tfrac12\cdot\tfrac12\int\frac{du}{u}=\tfrac14\ln u=\tfrac14\ln(x^{2}+1)+C_4.$$

Collecting all pieces and absorbing the constants into a single $$C$$, the required antiderivative is

$$\boxed{\displaystyle\int\frac{dx}{x(x^{4}-1)}=-\ln|x|+\tfrac14\ln|x-1|+\tfrac14\ln|x+1|+\tfrac14\ln(x^{2}+1)+C}. $$

(Any equivalent form obtained by combining logarithms is also correct.)

Objective: Evaluate $$I=\displaystyle\int \frac{1}{e^{x}-1}\,dx.$$

Step 1: Substitution. Put $$e^{x}=t$$, so $$dt=e^{x}\,dx \Rightarrow dx=\dfrac{dt}{t}.$$

Step 2: Rewrite the integrand.

$$I=\int \frac{1}{t-1}\cdot\frac{dt}{t}=\int\frac{dt}{t(t-1)}.$$

Step 3: Partial-fraction decomposition.

$$\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}.$$

Multiplying both sides by $$t(t-1)$$: $$1=A(t-1)+Bt.$$ Substitute roots:

• Set $$t=0$$: $$1 = -A \Rightarrow A = -1.$$
• Set $$t=1$$: $$1 = B \Rightarrow B = 1.$$

Hence

$$\frac{1}{t(t-1)}=-\frac{1}{t}+\frac{1}{t-1}.$$

Step 4: Integrate term-wise.

$$I=-\ln|t|+\ln|t-1|+C=\ln\!\left|\frac{t-1}{t}\right|+C.$$

Step 5: Back-substitution. Since $$t=e^{x}>0$$,

$$I=\ln\!\left|\frac{e^{x}-1}{e^{x}}\right|+C=\ln|e^{x}-1|-x+C.$$

Objective: Evaluate the integral $$I=\displaystyle\int \dfrac{x\,dx}{(x-1)(x-2)}.$$

1. Decompose into partial fractions

Assume

$$\dfrac{x}{(x-1)(x-2)}=\dfrac{A}{x-1}+\dfrac{B}{x-2}.$$

Multiply both sides by $$(x-1)(x-2)$$:

$$x=A(x-2)+B(x-1).$$

Expand the right-hand side:

$$x=A x-2A+B x-B=(A+B)x-(2A+B).$$

Because this must equal $$x=1\cdot x+0,$$ equate coefficients:

• Coefficient of $$x$$: $$A+B=1.$$
• Constant term: $$-(2A+B)=0\;\Rightarrow\;2A+B=0.$$

Solve the linear system:

From $$2A+B=0\;\Rightarrow\;B=-2A.$$ Substitute in $$A+B=1$$:

$$A-2A=1\;\Rightarrow\;-A=1\;\Rightarrow\;A=-1.$$

Then $$B=-2(-1)=2.$$

Hence

$$\dfrac{x}{(x-1)(x-2)}=\dfrac{-1}{x-1}+\dfrac{2}{x-2}.$$

2. Integrate term by term

$$I=\int\left(\frac{-1}{x-1}+\frac{2}{x-2}\right)dx=-\int\frac{dx}{x-1}+2\int\frac{dx}{x-2}.$$

Since $$\int \dfrac{dx}{x-c}=\log|x-c|+C,$$ we get

$$I=-\log|x-1|+2\log|x-2|+C.$$

3. Combine logarithms

Use $$a\log p-b\log q=\log p^{a}-\log q^{b}=\log\dfrac{p^{a}}{q^{b}}.$$

$$I=2\log|x-2|-\log|x-1|+C=\log|x-2|^{2}-\log|x-1|+C=\log\left|\dfrac{(x-2)^{2}}{x-1}\right|+C.$$

4. Choose the correct option

This matches option (B).

We have to evaluate the integral

$$\int \frac{dx}{x\,(x^{2}+1)}.$$

1. Resolve into partial fractions.

$$\frac{1}{x\,(x^{2}+1)} = \frac{A}{x} + \frac{Bx + C}{x^{2}+1}.$$

Multiply both sides by $$x\,(x^{2}+1)$$:

$$1 = A\,(x^{2}+1) + (Bx + C)\,x = (A+B)x^{2} + Cx + A.$$

Equating coefficients with $$1 = 0\,x^{2} + 0\,x + 1$$:

$$A+B = 0,\qquad C = 0,\qquad A = 1.$$

Hence $$A = 1,\;B = -1,\;C = 0$$, so

$$\frac{1}{x\,(x^{2}+1)} = \frac{1}{x} - \frac{x}{x^{2}+1}.$$

2. Integrate term-wise.

$$\int \frac{dx}{x\,(x^{2}+1)} = \int \frac{dx}{x} - \int \frac{x\,dx}{x^{2}+1}.$$

(i) $$\displaystyle \int \frac{dx}{x} = \log|x|.$$

(ii) For $$\displaystyle \int \frac{x\,dx}{x^{2}+1}$$, put $$u = x^{2}+1$$, so $$du = 2x\,dx$$, giving $$\tfrac12\log(x^{2}+1).$$

Therefore

$$\int \frac{dx}{x\,(x^{2}+1)} = \log|x| - \tfrac12\log(x^{2}+1) + C.$$

3. Choose the correct option. The result matches option (A).

We integrate by parts.

Let $$I = \int x \cos x\,dx$$.

Choose
$$u = x \;\;\Longrightarrow\;\; du = dx,$$
$$dv = \cos x\,dx \;\;\Longrightarrow\;\; v = \sin x.$$

By the integration-by-parts formula $$\int u\,dv = uv-\int v\,du$$, we obtain

$$I = x\sin x - \int \sin x\,dx.$$

Since $$\int \sin x\,dx = -\cos x + C_1$$ (where $$C_1$$ is a constant of integration),

$$I = x\sin x - \bigl(-\cos x\bigr) + C_1 = x\sin x + \cos x + C.$$

Thus,

$$\displaystyle \int x \cos x\,dx = x\sin x + \cos x + C.$$

Let $$I = \displaystyle\int \log x\,dx$$. We treat $$\log x$$ as the natural logarithm $$\ln x$$.

Apply integration by parts, choosing

• $$u = \log x \;\;\Longrightarrow\;\; du = \dfrac{1}{x}\,dx$$
• $$dv = dx \;\;\Longrightarrow\;\; v = x$$

Then the integration-by-parts formula $$\int u\,dv = uv - \int v\,du$$ gives

$$I = x\,\log x \;\; - \displaystyle\int x \cdot \dfrac{1}{x}\,dx$$

Simplify the integrand in the remaining integral:

$$I = x\,\log x - \int 1\,dx$$

Since $$\int 1\,dx = x$$, we obtain

$$I = x\,\log x - x + C$$,

where $$C$$ is the constant of integration.

Goal: Evaluate $$\int x\,e^x\,dx$$.

Method – Integration by parts
Recall the rule
$$\int u\,dv = u\,v-\int v\,du,$$ where $$u$$ and $$dv$$ are chosen functions of the variable of integration.

1. Choose
   $$u = x \quad\Longrightarrow\quad du = dx,$$
   $$dv = e^x\,dx \quad\Longrightarrow\quad v = \int e^x\,dx = e^x.$$

2. Apply the formula
   $$\int x\,e^x\,dx = u\,v - \int v\,du$$ $$\qquad\qquad\ = x\,e^x - \int e^x\,dx.$$

3. Integrate the remaining integral
   $$\int e^x\,dx = e^x,$$ so

   $$x\,e^x - \int e^x\,dx = x\,e^x - e^x + C.$$ (“+ C” is the constant of integration.)

4. Simplify
   Factor out $$e^x$$:

   $$x\,e^x - e^x + C = (x-1)\,e^x + C.$$

Result:
$$\displaystyle \int x\,e^x\,dx = (x-1)\,e^x + C.$$

We have to evaluate the integral

$$I = \int \dfrac{x\,\sin^{-1}x}{\sqrt{1 - x^2}}\,dx$$

Step 1 – Substitution
Put $$x = \sin u$$. Then $$u = \sin^{-1}x$$ and

$$dx = \cos u\,du, \qquad \sqrt{1 - x^2} = \sqrt{1 - \sin^{2}u} = \cos u.$$

Substituting in the integral:

$$I = \int \dfrac{(\sin u)(u)}{\cos u}\,(\cos u\,du) = \int u\sin u\,du.$$

Step 2 – Integration by parts
Choose:
  First function $$f(u)=u\;(\text{easily differentiable}),$$
  Second function $$g'(u)=\sin u\,(\text{easily integrable}).$$

Then $$g(u)=\int \sin u\,du = -\cos u.$$

By the integration‐by‐parts formula, $$\int f\,g'\,du = f\,g - \int f'\,g\,du$$:

$$\int u\sin u\,du = u(-\cos u) - \int 1\,(-\cos u)\,du$$

$$\phantom{\int u\sin u\,du} = -u\cos u + \int \cos u\,du$$

$$\phantom{\int u\sin u\,du} = -u\cos u + \sin u + C.$$

Step 3 – Back-substitution
Recall $$u = \sin^{-1}x$$, $$\cos u = \sqrt{1 - x^2}$$ and $$\sin u = x.$$ Hence

$$I = -\big(\sin^{-1}x\big)\sqrt{1 - x^2} + x + C.$$

Thus,

$$\boxed{\displaystyle\int \dfrac{x\,\sin^{-1}x}{\sqrt{1 - x^2}}\,dx = x - (\sin^{-1}x)\sqrt{1 - x^2} + C}.$$

We have to evaluate the indefinite integral

$$\displaystyle\int e^x \sin x\,dx$$

This integrand is a product of two elementary functions, so we use integration by parts twice.

Step 1 : First application of integration by parts

Choose

• $$u = \sin x \;\;\;\Longrightarrow\;\;\;du = \cos x\,dx$$
• $$dv = e^x\,dx \;\;\;\Longrightarrow\;\;\;v = e^x$$

By the rule $$\int u\,dv = uv-\int v\,du$$ we get

$$\int e^x \sin x\,dx = e^x \sin x - \int e^x \cos x\,dx$$

Step 2 : Second application of integration by parts to $$\int e^x \cos x\,dx$$

Now integrate $$\int e^x \cos x\,dx$$ in the same way. Choose

• $$u = \cos x \;\;\;\Longrightarrow\;\;\;du = -\sin x\,dx$$
• $$dv = e^x\,dx \;\;\;\Longrightarrow\;\;\;v = e^x$$

Thus

$$\int e^x \cos x\,dx = e^x \cos x - \int e^x(-\sin x)\,dx = e^x \cos x + \int e^x \sin x\,dx$$

Step 3 : Forming an equation for the required integral

Let $$I = \int e^x \sin x\,dx$$. Substituting the expression obtained in Step 2 into the result of Step 1 gives

$$I = e^x \sin x - \bigl(e^x \cos x + I\bigr)$$

Simplify the right–hand side:

$$I = e^x \sin x - e^x \cos x - I$$

Bring the term $$-I$$ to the left:

$$I + I = e^x \sin x - e^x \cos x$$

$$2I = e^x(\sin x - \cos x)$$

Step 4 : Solve for $$I$$

$$I = \dfrac{1}{2}e^x(\sin x - \cos x) + C$$

where $$C$$ is the constant of integration.

Verification (optional)

Differentiating $$\frac12 e^x(\sin x - \cos x)$$ indeed yields $$e^x \sin x$$, confirming the result.

Therefore,

$$\boxed{\displaystyle\int e^x \sin x\,dx = \frac{1}{2}e^x(\sin x - \cos x) + C}$$

We want to evaluate the indefinite integral

$$\displaystyle \int x \sin x \, dx.$$

The usual technique here is integration by parts, whose rule is

$$\displaystyle \int u \, dv = u v - \int v \, du.$$

Choose

• $$u = x \;\;\Rightarrow\;\; du = dx,$$
• $$dv = \sin x \, dx \;\;\Rightarrow\;\; v = -\cos x,$$ because $$\dfrac{d}{dx}(-\cos x)=\sin x.$$

Substituting these into the integration-by-parts formula gives

$$\int x \sin x \, dx = u v - \int v \, du = x\,( -\cos x) \, - \int ( -\cos x)\,(dx).$$

Simplify the right-hand side:

$$= -x \cos x + \int \cos x \, dx.$$

The remaining integral is elementary:

$$\int \cos x \, dx = \sin x + C_1,$$ where $$C_1$$ is a constant of integration.

Putting it back,

$$\int x \sin x \, dx = -x \cos x + \sin x + C_1.$$

Renaming the constant $$C_1$$ simply as $$C,$$ the final antiderivative is

$$\boxed{\int x \sin x \, dx = -x \cos x + \sin x + C}.$$

We have to evaluate the indefinite integral $$\int x \sin 3x\,dx$$.

Step 1 : Choose parts for integration by parts

• Take $$u = x$$ (because differentiating $$x$$ simplifies it).
• Take $$dv = \sin 3x\,dx$$.

Differentiate and integrate the chosen parts:

• $$du = dx.$$
• To find $$v$$, integrate $$dv$$:
    $$v = \int \sin 3x\,dx = -\frac{\cos 3x}{3}.$$

Step 2 : Apply integration by parts

The formula $$\int u\,dv = u v - \int v\,du$$ gives

$$\int x \sin 3x\,dx = x\left(-\frac{\cos 3x}{3}\right) - \int\left(-\frac{\cos 3x}{3}\right)dx.$$

Simplify the right-hand side:

$$= -\frac{x\cos 3x}{3} + \frac{1}{3}\int \cos 3x\,dx.$$

Step 3 : Integrate the remaining integral

Evaluate $$\int \cos 3x\,dx$$:

$$\int \cos 3x\,dx = \frac{\sin 3x}{3}.$$

Substitute this result:

$$-\frac{x\cos 3x}{3} + \frac{1}{3}\left(\frac{\sin 3x}{3}\right) + C.$$

Step 4 : Final simplification

$$-\frac{x\cos 3x}{3} + \frac{\sin 3x}{9} + C.$$

Thus, the required antiderivative is $$-\dfrac{x\cos 3x}{3} + \dfrac{\sin 3x}{9} + C$$.

We need to evaluate the indefinite integral

$$\displaystyle \int x^2 e^x \,dx$$

The integrand is a product of a polynomial $$x^2$$ and an exponential function $$e^x$$, so we apply integration by parts.

Step 1: First application of integration by parts

Recall the formula $$\int u\,dv = u\,v - \int v\,du$$.

Choose

• $$u = x^2 \;\;\Rightarrow\;\; du = 2x\,dx$$
• $$dv = e^x\,dx \;\;\Rightarrow\;\; v = e^x$$  (because $$\int e^x\,dx = e^x$$)

Then

$$\int x^2 e^x\,dx = x^2 e^x - \int e^x (2x)\,dx$$

Simplify the remaining integral:

$$\int x^2 e^x\,dx = x^2 e^x - 2 \int x e^x\,dx$$

Step 2: Second application of integration by parts

Now integrate $$\int x e^x\,dx$$ in the same manner.

Let

• $$u_1 = x \;\;\Rightarrow\;\; du_1 = dx$$
• $$dv_1 = e^x\,dx \;\;\Rightarrow\;\; v_1 = e^x$$

Hence

$$\int x e^x\,dx = x e^x - \int e^x\,dx$$

But $$\int e^x\,dx = e^x$$, so

$$\int x e^x\,dx = x e^x - e^x$$

Step 3: Substitute back

Return to the result from Step 1:

$$\int x^2 e^x\,dx = x^2 e^x - 2\bigl( x e^x - e^x \bigr)$$

Distribute the factor $$-2$$ inside the brackets:

$$\int x^2 e^x\,dx = x^2 e^x - 2x e^x + 2 e^x$$

Step 4: Factor and add the constant of integration

All three terms contain $$e^x$$, so we factor it:

$$\int x^2 e^x\,dx = e^x\,(x^2 - 2x + 2) + C$$

where $$C$$ is the constant of integration.

Therefore, the required integral is

$$\boxed{\int x^2 e^x\,dx = e^x\,(x^2 - 2x + 2) + C}$$

Integral to evaluate: $$\int x\log x\,dx$$ (with $$x>0$$ so that $$\log x$$ is defined).

Step 1: Choose functions for integration by parts

• Let $$u = \log x$$ (because its derivative is simple).
• Let $$dv = x\,dx$$ (easy to integrate).

Step 2: Differentiate $$u$$ and integrate $$dv$$

• $$du = \dfrac{1}{x}\,dx$$
• $$v = \int x\,dx = \dfrac{x^{2}}{2}$$

Step 3: Apply the formula $$\int u\,dv = u\,v \; - \int v\,du$$

$$\int x\log x\,dx = \log x\,\cdot\,\dfrac{x^{2}}{2} - \int \dfrac{x^{2}}{2}\,\cdot\,\dfrac{1}{x}\,dx$$

Step 4: Simplify the remaining integral

Inside the integral, $$\dfrac{x^{2}}{2}\times\dfrac{1}{x}=\dfrac{x}{2}$$. Therefore

$$\int x\log x\,dx = \dfrac{x^{2}}{2}\log x - \dfrac{1}{2}\int x\,dx$$

Step 5: Evaluate $$\int x\,dx$$

$$\int x\,dx = \dfrac{x^{2}}{2}$$

Step 6: Substitute and tidy up

$$\int x\log x\,dx = \dfrac{x^{2}}{2}\log x - \dfrac{1}{2}\cdot\dfrac{x^{2}}{2} + C$$

$$\int x\log x\,dx = \dfrac{x^{2}}{2}\log x - \dfrac{x^{2}}{4} + C$$

Final result: $$\boxed{\dfrac{x^{2}}{2}\log x - \dfrac{x^{2}}{4} + C}$$

We have to evaluate the indefinite integral

$$\displaystyle I = \int x \log 2x\, \mathrm{d}x$$

Method 1: Integration by parts

Choose

• $$u = \log 2x$$   so that  $$\displaystyle \mathrm{d}u = \frac{1}{x}\, \mathrm{d}x$$
• $$\mathrm{d}v = x\, \mathrm{d}x$$   so that  $$v = \frac{x^{2}}{2}$$

Applying the by-parts formula $$\displaystyle \int u\,\mathrm{d}v = uv - \int v\,\mathrm{d}u$$, we get

$$I = \frac{x^{2}}{2}\,\log 2x\; - \; \int \frac{x^{2}}{2}\, \frac{1}{x}\, \mathrm{d}x$$

Simplify the remaining integral:

$$\int \frac{x^{2}}{2}\, \frac{1}{x}\, \mathrm{d}x = \frac12 \int x\, \mathrm{d}x = \frac12 \cdot \frac{x^{2}}{2} = \frac{x^{2}}{4}$$

Therefore,

$$I = \frac{x^{2}}{2}\,\log 2x - \frac{x^{2}}{4} + C$$

Method 2: Expanding the logarithm (check)

Using $$\log 2x = \log 2 + \log x$$,

$$I = \int x(\log 2 + \log x)\, \mathrm{d}x = \log 2 \int x\, \mathrm{d}x + \int x \log x\, \mathrm{d}x$$

The second integral is a standard result $$\int x \log x\, \mathrm{d}x = \frac{x^{2}}{2}\log x - \frac{x^{2}}{4}$$. Substituting, one again obtains

$$I = \frac{x^{2}}{2}\log 2x - \frac{x^{2}}{4} + C$$

Both approaches agree, so the required antiderivative is

$$\boxed{\displaystyle \int x \log 2x\, \mathrm{d}x = \frac{x^{2}}{2}\log(2x) - \frac{x^{2}}{4} + C}$$

We want to evaluate the indefinite integral

$$\int x^2 \log x\,dx,$$

where $$x>0$$ so that $$\log x$$ is defined.

Step 1: Choose the functions for integration by parts

• Take the logarithmic part as the first function:
  $$u = \log x \;\;\Longrightarrow\;\; du = \frac{1}{x}\,dx.$$
• The remaining factor is the second function to be integrated:
  $$dv = x^2\,dx \;\;\Longrightarrow\;\; v = \int x^2\,dx = \frac{x^3}{3}.$$

Step 2: Apply the integration-by-parts formula

The formula is $$\int u\,dv = uv - \int v\,du.$$ Substituting the chosen $$u$$ and $$v$$ gives

$$\int x^2 \log x\,dx = \left(\log x\right)\left(\frac{x^3}{3}\right) - \int \frac{x^3}{3}\cdot\frac{1}{x}\,dx.$$

Step 3: Simplify the remaining integral

Inside the new integral, combine the powers of $$x$$:

$$\int \frac{x^3}{3}\cdot\frac{1}{x}\,dx = \frac{1}{3}\int x^{3-1}\,dx = \frac{1}{3}\int x^2\,dx.$$

Now integrate $$x^2$$:

$$\frac{1}{3}\int x^2\,dx = \frac{1}{3}\left(\frac{x^3}{3}\right) = \frac{x^3}{9}.$$

Step 4: Assemble the result and add the constant of integration

Putting everything together,

$$\int x^2 \log x\,dx = \frac{x^3}{3}\,\log x - \frac{x^3}{9} + C.$$

Hence the required indefinite integral is

$$\boxed{\displaystyle \int x^2 \log x\,dx = \frac{x^3}{3}\,\log x - \frac{x^3}{9} + C},$$

valid for $$x>0$$.

Let $$I = \int x \sin^{-1}x\,dx$$.

We use integration by parts, choosing
$$u = \sin^{-1}x \;\;\;\;\;\; dv = x\,dx.$$

Then
$$du = \frac{1}{\sqrt{1-x^{2}}}\,dx,\qquad v = \int x\,dx = \frac{x^{2}}{2}.$$

By the formula $$\int u\,dv = uv - \int v\,du$$ we get

$$I = \frac{x^{2}}{2}\sin^{-1}x\; - \int \frac{x^{2}}{2}\,\frac{dx}{\sqrt{1-x^{2}}}.$$

Denote the remaining integral by $$J$$:

$$J = \frac{1}{2}\int \frac{x^{2}}{\sqrt{1-x^{2}}}\,dx.$$

Rewrite the numerator: $$x^{2} = 1 - (1-x^{2}).$$ Hence

$$J = \frac{1}{2}\int \left[\frac{1}{\sqrt{1-x^{2}}} - \sqrt{1-x^{2}}\right] dx$$
$$\phantom{J}= \frac{1}{2}\Bigg[\int \frac{dx}{\sqrt{1-x^{2}}} \;\; - \;\; \int \sqrt{1-x^{2}}\,dx\Bigg].$$

• First part: $$\displaystyle \int \frac{dx}{\sqrt{1-x^{2}}} = \sin^{-1}x + C.$$
• Second part: To evaluate $$\int \sqrt{1-x^{2}}\,dx$$ set $$x = \sin\theta\;( -\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}),\; dx = \cos\theta\,d\theta.$$ Then
  $$\sqrt{1-x^{2}} = \cos\theta,$$
  so
  $$\int \sqrt{1-x^{2}}\,dx = \int \cos^{2}\theta\,d\theta.$$
  Using $$\cos^{2}\theta = \tfrac{1+\cos2\theta}{2}$$,
  $$\int \cos^{2}\theta\,d\theta = \int \frac{1+\cos2\theta}{2}\,d\theta = \frac{\theta}{2} + \frac{\sin2\theta}{4} + C.$$
  Because $$\theta = \sin^{-1}x$$ and $$\sin2\theta = 2\sin\theta\cos\theta = 2x\sqrt{1-x^{2}},$$ we obtain
  $$\int \sqrt{1-x^{2}}\,dx = \frac{1}{2}\sin^{-1}x + \frac{x\sqrt{1-x^{2}}}{2} + C.$$

Putting the two parts together,

$$\begin{aligned} J &= \frac{1}{2}\Bigg[\sin^{-1}x - \Big(\tfrac{1}{2}\sin^{-1}x + \tfrac{x\sqrt{1-x^{2}}}{2}\Big)\Bigg]\\[4pt] &= \frac{1}{2}\Bigg[\frac{1}{2}\sin^{-1}x - \frac{x\sqrt{1-x^{2}}}{2}\Bigg]\\[4pt] &= \frac{1}{4}\sin^{-1}x - \frac{1}{4}x\sqrt{1-x^{2}}. \end{aligned}$$

Return to $$I$$:

$$\begin{aligned} I &= \frac{x^{2}}{2}\sin^{-1}x - J\\[4pt] &= \frac{x^{2}}{2}\sin^{-1}x - \Big(\tfrac{1}{4}\sin^{-1}x - \tfrac{1}{4}x\sqrt{1-x^{2}}\Big)\\[4pt] &= \Big(\tfrac{x^{2}}{2} - \tfrac{1}{4}\Big)\sin^{-1}x + \tfrac{1}{4}x\sqrt{1-x^{2}} + C. \end{aligned}$$

Combine the coefficients: $$\tfrac{x^{2}}{2} - \tfrac{1}{4} = \tfrac{2x^{2} - 1}{4}.$$

Therefore

$$\boxed{\displaystyle \int x \sin^{-1}x\,dx = \frac{2x^{2}-1}{4}\,\sin^{-1}x + \frac{x}{4}\sqrt{1-x^{2}} + C}.$$

Let $$I = \int x \tan^{-1}x \, dx$$.

Apply integration by parts, $$\int u\,dv = u v - \int v\,du$$.

Choose $$u = \tan^{-1}x \;\Rightarrow\; du = \dfrac{1}{1+x^{2}}\,dx$$
and $$dv = x\,dx \;\Rightarrow\; v = \dfrac{x^{2}}{2}.$$

Then

$$I = \frac{x^{2}}{2}\tan^{-1}x - \int \frac{x^{2}}{2}\cdot\frac{1}{1+x^{2}}\,dx.$$

Rewrite the integrand:

$$\frac{x^{2}}{1+x^{2}} = 1 - \frac{1}{1+x^{2}}.$$

Hence

$$I = \frac{x^{2}}{2}\tan^{-1}x - \frac{1}{2}\int\left(1 - \frac{1}{1+x^{2}}\right)dx.$$

Integrate term by term:

$$\int 1\,dx = x, \qquad \int \frac{1}{1+x^{2}}\,dx = \tan^{-1}x.$$

Therefore

$$I = \frac{x^{2}}{2}\tan^{-1}x - \frac{1}{2}(x - \tan^{-1}x) + C.$$

Simplify:

$$I = \frac{x^{2}}{2}\tan^{-1}x - \frac{x}{2} + \frac{1}{2}\tan^{-1}x + C$$

$$I = \frac{1}{2}(x^{2}+1)\tan^{-1}x - \frac{x}{2} + C.$$

Thus, $$\int x \tan^{-1}x\,dx = \frac{1}{2}(x^{2}+1)\tan^{-1}x - \frac{x}{2} + C.$$

We have to evaluate the indefinite integral

$$I=\int x\,\cos^{-1}x\,dx\qquad(-1\le x\le 1).$$

Step 1: Integration by parts. Choose $$u=\cos^{-1}x$$ and $$dv = x\,dx$$, so $$du = -\dfrac{1}{\sqrt{1-x^{2}}}\,dx$$ and $$v = \dfrac{x^{2}}{2}.$$

Then

$$I = \frac{x^{2}}{2}\,\cos^{-1}x + \frac{1}{2}\int \frac{x^{2}}{\sqrt{1-x^{2}}}\,dx. \qquad (1)$$

Step 2: Simplify the remaining integral. Write $$x^{2}=1-(1-x^{2})$$:

$$\frac{x^{2}}{\sqrt{1-x^{2}}}=\frac{1}{\sqrt{1-x^{2}}}-\sqrt{1-x^{2}}.$$

Hence

$$\frac{1}{2}\int \frac{x^{2}}{\sqrt{1-x^{2}}}\,dx = \frac{1}{2}\left[\int\frac{dx}{\sqrt{1-x^{2}}} - \int \sqrt{1-x^{2}}\,dx\right].$$

Use the two standard results $$\int \dfrac{dx}{\sqrt{1-x^{2}}}=\sin^{-1}x$$ and $$\int\sqrt{1-x^{2}}\,dx = \dfrac{x}{2}\sqrt{1-x^{2}}+\dfrac{1}{2}\sin^{-1}x.$$

Substituting,

$$\frac{1}{2}\int \frac{x^{2}}{\sqrt{1-x^{2}}}\,dx = \frac{1}{2}\sin^{-1}x - \frac{1}{2}\left[\frac{x}{2}\sqrt{1-x^{2}}+\frac{1}{2}\sin^{-1}x\right] = \frac{1}{4}\sin^{-1}x - \frac{x}{4}\sqrt{1-x^{2}}. \qquad (2)$$

Step 3: Combine. Substituting (2) into (1):

$$I = \frac{x^{2}}{2}\,\cos^{-1}x + \frac{1}{4}\sin^{-1}x - \frac{x}{4}\sqrt{1-x^{2}} + C.$$

Using $$\sin^{-1}x = \dfrac{\pi}{2}-\cos^{-1}x$$, the term $$\tfrac14\sin^{-1}x$$ becomes $$\tfrac{\pi}{8} - \tfrac14\cos^{-1}x$$. The constant $$\tfrac{\pi}{8}$$ combines with $$C$$ into a new constant (also written $$C$$). Then

$$I = \frac{x^{2}}{2}\cos^{-1}x - \frac{1}{4}\cos^{-1}x - \frac{x}{4}\sqrt{1-x^{2}} + C = \frac{(2x^{2}-1)\cos^{-1}x - x\sqrt{1-x^{2}}}{4}+C.$$

Let
$$I = \displaystyle\int (\sin^{-1} x)^2\,dx$$

Step 1 — Integration by parts
Choose $$u = (\sin^{-1}x)^2$$ and $$dv = dx$$.
Then $$du = 2\sin^{-1}x\,\dfrac{1}{\sqrt{1-x^2}}\,dx$$ and $$v = x$$.

Applying $$\displaystyle\int u\,dv = uv-\int v\,du$$ we get
$$I = x(\sin^{-1}x)^2 - 2\int \dfrac{x\,\sin^{-1}x}{\sqrt{1-x^2}}\,dx$$

Denote
$$J = \int \dfrac{x\,\sin^{-1}x}{\sqrt{1-x^2}}\,dx$$

Step 2 — Simplify $$J$$ by the substitution $$\theta = \sin^{-1}x$$
Then $$x = \sin\theta,\;dx = \cos\theta\,d\theta,\;\sqrt{1-x^2}=\cos\theta$$.

Thus
$$J = \int \dfrac{\sin\theta\,\theta}{\cos\theta}\,(\cos\theta\,d\theta) = \int \theta\sin\theta\,d\theta$$

Step 3 — Evaluate $$\displaystyle\int \theta\sin\theta\,d\theta$$
Use integration by parts again with $$a = \theta$$, $$db = \sin\theta\,d\theta$$.
Then $$da = d\theta$$ and $$b = -\cos\theta$$.

Hence
$$\int \theta\sin\theta\,d\theta = -\theta\cos\theta + \int \cos\theta\,d\theta = -\theta\cos\theta + \sin\theta + C$$

Discard the constant temporarily because it will be absorbed later. Therefore
$$J = -\theta\cos\theta + \sin\theta$$

Return to $$x$$ using $$\theta = \sin^{-1}x$$:
$$J = -\sin^{-1}x\,\sqrt{1-x^2} + x$$

Step 4 — Assemble the result for $$I$$
Recall $$I = x(\sin^{-1}x)^2 - 2J$$, so

$$\begin{aligned} I &= x(\sin^{-1}x)^2 - 2\bigl[-\sin^{-1}x\,\sqrt{1-x^2} + x\bigr] \\[4pt] &= x(\sin^{-1}x)^2 + 2\sin^{-1}x\,\sqrt{1-x^2} - 2x + C. \end{aligned}$$

Therefore
$$\boxed{\displaystyle\int (\sin^{-1}x)^2\,dx = x(\sin^{-1}x)^2 + 2\sin^{-1}x\,\sqrt{1 - x^2} - 2x + C}$$

Let

$$I = \int \dfrac{x\,\cos^{-1}x}{\sqrt{1-x^{2}}}\,dx$$

Put $$x = \cos\theta$$. Then $$dx = -\sin\theta\,d\theta$$ and $$\sqrt{1-x^{2}} = \sqrt{1-\cos^{2}\theta}=\sin\theta\;\;(0<\theta<\pi).$$

Substituting,

$$I = \int \dfrac{\cos\theta\,(\cos^{-1}x)}{\sin\theta}\;(-\sin\theta\,d\theta)$$

but $$\cos^{-1}x = \theta$$, so

$$I = -\int \theta\cos\theta\,d\theta.$$

Integrate by parts: take $$u=\theta\;(du = d\theta)$$ and $$dv = \cos\theta\,d\theta\;(v = \sin\theta).$$

Then

$$\int \theta\cos\theta\,d\theta = \theta\sin\theta - \int \sin\theta\,d\theta = \theta\sin\theta + \cos\theta + C.$$

Hence

$$I = -\bigl(\theta\sin\theta + \cos\theta\bigr) + C.$$

Return to $$x$$: $$\theta = \cos^{-1}x,$$ $$\sin\theta = \sqrt{1-x^{2}},$$ and $$\cos\theta = x.$$ Thus

$$I = -\bigl(\cos^{-1}x\;\sqrt{1-x^{2}} + x\bigr) + C.$$

Therefore

$$\boxed{\displaystyle\int \dfrac{x\,\cos^{-1}x}{\sqrt{1-x^{2}}}\,dx = -\sqrt{1-x^{2}}\,\cos^{-1}x\; - x + C}. $$

We have to evaluate the indefinite integral

$$\int x \sec^2 x\,dx$$

This is a product of an algebraic factor $$x$$ and a trigonometric factor $$\sec^2 x$$, so we use integration by parts.

Step 1 – Choose u and v':
Take $$u = x$$ (easy to differentiate) and $$dv = \sec^2 x\,dx$$ (easy to integrate).

Then
$$du = dx$$
$$v = \int \sec^2 x\,dx = \tan x$$

Step 2 – Apply the formula:
For any two functions $$u$$ and $$v$$, $$\int u\,dv = uv-\int v\,du$$. Hence

$$\int x\sec^2 x\,dx = x\tan x - \int \tan x\,dx$$

Step 3 – Integrate $$\tan x$$:
Recall $$\int \tan x\,dx = -\ln|\cos x| + C$$ (one may verify by differentiating).

Substitute this result:

$$\int x\sec^2 x\,dx = x\tan x - \left(-\ln|\cos x|\right) + C$$

Step 4 – Simplify:
$$x\tan x - (-\ln|\cos x|) = x\tan x + \ln|\cos x|$$

Therefore

$$\boxed{\displaystyle \int x\sec^2 x\,dx = x\tan x + \ln|\cos x| + C}$$

(An equivalent answer is $$x\tan x - \ln|\sec x| + C$$ since $$\ln|\cos x| = -\ln|\sec x|$$.)

Goal: Evaluate the indefinite integral $$\displaystyle\int \tan^{-1}x\,dx$$.

Step 1 – Choose the parts for integration by parts.
Recall the rule $$\displaystyle\int u\,dv = uv-\int v\,du$$.
A useful heuristic (ILATE) suggests taking an inverse trigonometric function as u and an algebraic function as dv. Hence let

• $$u = \tan^{-1}x$$
• $$dv = dx$$

Step 2 – Differentiate u and integrate dv.

• Derivative: $$du = \dfrac{1}{1+x^2}\,dx$$  because $$\dfrac{d}{dx}(\tan^{-1}x)=\dfrac{1}{1+x^2}$$.
• Integral: $$v = \int dx = x$$.

Step 3 – Apply the integration-by-parts formula.

$$\int \tan^{-1}x\,dx = x\tan^{-1}x - \int x\left(\dfrac{1}{1+x^2}\right)dx$$

Step 4 – Evaluate the remaining integral.
Need to compute $$\displaystyle\int \dfrac{x}{1+x^2}\,dx$$.

Put $$t = 1 + x^2 \;\Rightarrow\; dt = 2x\,dx \;\Rightarrow\; x\,dx = \dfrac{dt}{2}$$.

Substitution converts the integral to

$$\int \frac{x}{1+x^2}\,dx = \int \frac{1}{t}\left(\frac{dt}{2}\right) = \frac12\int \frac{dt}{t} = \frac12\ln|t| + C_1$$

Undo the substitution: $$t = 1 + x^2$$, so the primitive is $$\dfrac12\ln(1+x^2) + C_1$$.

Step 5 – Combine the results.

Insert this back into Step 3:

$$\int \tan^{-1}x\,dx = x\tan^{-1}x - \left(\frac12\ln(1+x^2)\right) + C$$

(The two arbitrary constants merge into a single constant $$C$$.)

Final result:

$$\boxed{\displaystyle\int \tan^{-1}x\,dx = x\tan^{-1}x - \frac12\ln(1+x^2) + C}$$

We have to evaluate the indefinite integral

$$I = \int x (\log x)^2\,dx.$$

Throughout, log denotes the natural logarithm.

Step 1: First application of integration by parts

Recall the rule $$\int u\,dv = u v - \int v\,du.$$

Choose

$$u = (\log x)^2,\qquad dv = x\,dx.$$

Then

$$du = 2\,(\log x)\,\frac{1}{x}\,dx = \frac{2\,\log x}{x}\,dx,\qquad v = \int x\,dx = \frac{x^2}{2}.$$

Substituting in the by-parts formula,

$$I = (\log x)^2\,\frac{x^2}{2} - \int \frac{x^2}{2}\,\frac{2\,\log x}{x}\,dx.$$

Simplify the integrand of the remaining integral:

$$I = \frac{x^2}{2}(\log x)^2 - \int x\,(\log x)\,dx.$$

Step 2: Second application of integration by parts to \(\displaystyle \int x\log x\,dx\)

Set

$$u_1 = \log x,\qquad dv_1 = x\,dx.$$

Then

$$du_1 = \frac{1}{x}\,dx,\qquad v_1 = \frac{x^2}{2}.$$

Hence

$$\int x\log x\,dx = \log x\,\frac{x^2}{2} - \int \frac{x^2}{2}\,\frac{1}{x}\,dx$$ $$\phantom{\int x\log x\,dx} = \frac{x^2}{2}\,\log x - \frac{1}{2}\int x\,dx.$$

The remaining integral is elementary:

$$\int x\,dx = \frac{x^2}{2},$$

so

$$\int x\log x\,dx = \frac{x^2}{2}\,\log x - \frac{1}{2}\cdot\frac{x^2}{2} = \frac{x^2}{2}\,\log x - \frac{x^2}{4}.$$

Step 3: Substitute back into \(I\)

Returning to the expression for I obtained in Step 1,

$$I = \frac{x^2}{2}(\log x)^2 - \Bigl[\frac{x^2}{2}\,\log x - \frac{x^2}{4}\Bigr] + C,$$

where $$C$$ is the constant of integration.

Simplify the terms:

$$I = \frac{x^2}{2}(\log x)^2 - \frac{x^2}{2}\,\log x + \frac{x^2}{4} + C.$$

This can also be written in a compact form:

$$I = \frac{x^2}{4}\bigl[2(\log x)^2 - 2\log x + 1\bigr] + C.$$

Thus, the required integral is obtained.

Let $$I = \int (x^2 + 1)\log x\,dx$$  (the integrand is defined for $$x > 0$$).

Step 1 – Choose the functions for integration by parts
Take $$u = \log x$$ (because its derivative is simpler) and $$dv = (x^2 + 1)\,dx$$.

Then
$$du = \frac{1}{x}\,dx$$
$$v = \int (x^2 + 1)\,dx = \frac{x^3}{3} + x$$

Step 2 – Apply the integration-by-parts formula
$$I = u\,v - \int v\,du$$

Substituting $$u, v, du$$:

$$I = \bigl(\tfrac{x^3}{3} + x\bigr)\log x - \int \left(\tfrac{x^3}{3} + x\right)\frac{1}{x}\,dx$$

Step 3 – Simplify the remaining integral
Inside the integral:
$$\left(\tfrac{x^3}{3} + x\right)\frac{1}{x} = \tfrac{x^2}{3} + 1$$

Thus
$$I = \bigl(\tfrac{x^3}{3} + x\bigr)\log x - \int \left(\tfrac{x^2}{3} + 1\right)dx$$

Step 4 – Integrate the simpler polynomial
$$\int \left(\tfrac{x^2}{3} + 1\right)dx = \tfrac{1}{3}\int x^2 dx + \int 1\,dx = \tfrac{1}{3}\cdot\tfrac{x^3}{3} + x = \tfrac{x^3}{9} + x$$

Step 5 – Write the final expression
$$I = \bigl(\tfrac{x^3}{3} + x\bigr)\log x - \left(\tfrac{x^3}{9} + x\right) + C$$

That is,

$$I = \left(\frac{x^3}{3} + x\right)\log x - \frac{x^3}{9} - x + C$$

Let $$I = \int e^x(\sin x + \cos x)\,dx$$.

Observe that the integrand is the derivative of the single product $$e^x \sin x$$. Indeed, by the product rule,

$$\dfrac{d}{dx}\big(e^x\sin x\big) = e^x\cos x + e^x\sin x = e^x(\sin x + \cos x).$$

Hence

$$I = e^x\sin x + C,$$

where $$C$$ is the constant of integration.

We have to evaluate the indefinite integral

$$\int \dfrac{x\,e^x}{(1+x)^2}\,dx$$

Instead of beginning with an arbitrary substitution, first look for a function whose derivative might reproduce the given integrand. An expression that almost matches the integrand is $$\dfrac{e^x}{1+x}$$ because its denominator already contains the factor $$(1+x)^2$$ after differentiation by the Quotient Rule.

Step 1 – Differentiate the “guess”

Let $$f(x)=\dfrac{e^x}{1+x}$$. Using the Quotient Rule,

$$\frac{d}{dx}\left(f(x)\right)=\frac{(1+x)\,\dfrac{d}{dx}(e^x)-e^x\,\dfrac{d}{dx}(1+x)}{(1+x)^2}.$$

Compute each derivative explicitly:

• $$\dfrac{d}{dx}(e^x)=e^x$$
• $$\dfrac{d}{dx}(1+x)=1$$

Substitute these into the quotient formula:

$$\frac{d}{dx}\left(\dfrac{e^x}{1+x}\right)=\frac{(1+x)\,e^x-e^x\cdot1}{(1+x)^2}$$

Simplify the numerator:

$$=(1+x)e^x-e^x=e^x(1+x-1)=e^x\,x.$$

Hence

$$\frac{d}{dx}\left(\dfrac{e^x}{1+x}\right)=\dfrac{x\,e^x}{(1+x)^2}.$$

Step 2 – Recognise the exact derivative

The right–hand side is exactly the integrand in our problem. Therefore the antiderivative is immediately identified:

$$\int\dfrac{x\,e^x}{(1+x)^2}\,dx=\dfrac{e^x}{1+x}+C,$$

where $$C$$ is the required constant of integration.

Verification (optional)

Differentiate the obtained result to confirm:

$$\frac{d}{dx}\left(\dfrac{e^x}{1+x}\right)=\dfrac{x\,e^x}{(1+x)^2},$$

which matches the original integrand, completing the solution.

Let $$I=\displaystyle\int e^{x}\left(\dfrac{1+\sin x}{1+\cos x}\right)dx.$$

We look for a function $$g(x)$$ such that $$\dfrac{d}{dx}\bigl[e^{x}g(x)\bigr]=e^{x}\left(\dfrac{1+\sin x}{1+\cos x}\right).$$ Since $$\dfrac{d}{dx}\bigl[e^{x}g(x)\bigr]=e^{x}\bigl(g(x)+g'(x)\bigr),$$ we need

\[g(x)+g'(x)=\dfrac{1+\sin x}{1+\cos x}.\tag{1}\]

A convenient choice is $$g(x)=\dfrac{\sin x}{1+\cos x}.$$

Derivative of $$g(x)$$ — using the quotient rule,

$$g'(x)=\dfrac{\cos x\,(1+\cos x)-\sin x\,(-\sin x)}{(1+\cos x)^{2}}=\dfrac{\cos x+\cos^{2}x+\sin^{2}x}{(1+\cos x)^{2}}=\dfrac{1+\cos x}{(1+\cos x)^{2}}=\dfrac{1}{1+\cos x}.$$

Now

$$g(x)+g'(x)=\dfrac{\sin x}{1+\cos x}+\dfrac{1}{1+\cos x}=\dfrac{1+\sin x}{1+\cos x},$$

so condition (1) is satisfied.

Therefore

$$\dfrac{d}{dx}\Bigl[e^{x}\,\dfrac{\sin x}{1+\cos x}\Bigr]=e^{x}\left(\dfrac{1+\sin x}{1+\cos x}\right).$$

Integrating both sides,

$$\int e^{x}\left(\dfrac{1+\sin x}{1+\cos x}\right)dx=e^{x}\left(\dfrac{\sin x}{1+\cos x}\right)+C,$$

where $$C$$ is the constant of integration.

Hence the required integral is $$e^{x}\,\dfrac{\sin x}{1+\cos x}+C.$$

To integrate $$e^x\left(\dfrac{1}{x}-\dfrac{1}{x^2}\right)$$ we first look for a function whose derivative matches the integrand.

Take $$F(x)=\dfrac{e^x}{x}\;(x\neq 0).$$

Differentiating (treating $$e^x/x=e^x\,x^{-1}$$):

$$F'(x)=e^x\,x^{-1}+e^x\,(-x^{-2})=e^x\left(\dfrac{1}{x}-\dfrac{1}{x^2}\right).$$

This derivative is exactly the given integrand. Hence,

$$\int e^x\left(\dfrac{1}{x}-\dfrac{1}{x^2}\right)\,dx = F(x)+C = \dfrac{e^x}{x}+C,$$

where $$C$$ is the constant of integration and $$x\neq 0$$.

We have to evaluate the indefinite integral

$$\displaystyle \int \frac{(x-3)e^{x}}{(x-1)^3}\,dx.$$

The structure suggests that the integrand might be the derivative of a quotient involving the same exponential factor and a power of \((x-1)\). We verify this idea by direct differentiation.

Take the trial function

$$F(x)=\frac{e^{x}}{(x-1)^2}.$$

Differentiate it using the quotient (or product) rule. It is convenient to treat it as a product of two functions:

$$F(x)=e^{x}\,(x-1)^{-2}.$$

Let

• $$u=e^{x}\;\;\Longrightarrow\;\;u'=e^{x}$$
• $$v=(x-1)^{-2}\;\;\Longrightarrow\;\;v'=-2\,(x-1)^{-3}$$

Then, by the product rule,

$$F'(x)=u'v+uv' = e^{x}(x-1)^{-2}+e^{x}\bigl[-2\,(x-1)^{-3}\bigr].$$

Place the two terms over the common denominator \((x-1)^3\):

$$F'(x)=e^{x}\left[\,(x-1)\,-2\right](x-1)^{-3} =e^{x}\,(x-3)(x-1)^{-3}.$$

Re-introducing fractional notation yields

$$F'(x)=\frac{(x-3)e^{x}}{(x-1)^3},$$

which is exactly our integrand. Therefore

$$\int\frac{(x-3)e^{x}}{(x-1)^3}\,dx = F(x)+C = \frac{e^{x}}{(x-1)^2}+C,$$

where $$C$$ is the constant of integration.

We have to evaluate the indefinite integral

$$\displaystyle\int e^{2x}\,\sin x\,dx$$

Denote the required integral by

$$I = \int e^{2x}\,\sin x\,dx$$

Step 1 : First integration by parts

Choose  $$u = \sin x$$  and  $$dv = e^{2x}\,dx$$. Then

$$du = \cos x\,dx, \qquad v = \frac{e^{2x}}{2}$$

Using the formula $$\int u\,dv = u\,v-\int v\,du$$ we get

$$I = \sin x\,\frac{e^{2x}}{2}-\int \frac{e^{2x}}{2}\,\cos x\,dx$$

$$\Rightarrow\; I = \frac{e^{2x}\sin x}{2}-\frac12\int e^{2x}\cos x\,dx$$

Introduce a new symbol for the remaining integral:

$$J = \int e^{2x}\cos x\,dx$$

Thus

\[I = \frac{e^{2x}\sin x}{2}-\frac12\,J \tag{1}\]

Step 2 : Evaluate $$J$$ by another integration by parts

For $$J$$ take  $$u = \cos x$$  and  $$dv = e^{2x}\,dx$$, so

$$du = -\sin x\,dx, \qquad v = \frac{e^{2x}}{2}$$

Therefore

$$J = \cos x\,\frac{e^{2x}}{2}-\int \frac{e^{2x}}{2}\,(-\sin x)\,dx$$

$$\Rightarrow\; J = \frac{e^{2x}\cos x}{2}+\frac12\int e^{2x}\sin x\,dx$$

Noting that the integral on the right is again $$I$$, we have

\[J = \frac{e^{2x}\cos x}{2}+\frac12\,I \tag{2}\]

Step 3 : Solve the two equations for $$I$$

Substitute (2) into (1):

$$I = \frac{e^{2x}\sin x}{2}-\frac12\left(\frac{e^{2x}\cos x}{2}+\frac12 I\right)$$

Simplify term by term:

$$I = \frac{e^{2x}\sin x}{2}-\frac{e^{2x}\cos x}{4}-\frac14 I$$

Bring the $$\frac14 I$$ term to the left:

$$I + \frac14 I = \frac{e^{2x}\sin x}{2}-\frac{e^{2x}\cos x}{4}$$

$$\frac54 I = e^{2x}\left(\frac{\sin x}{2}-\frac{\cos x}{4}\right)$$

Step 4 : Isolate $$I$$

$$I = \frac45\,e^{2x}\left(\frac{\sin x}{2}-\frac{\cos x}{4}\right) + C$$

Distribute the factor $$\frac45$$ to obtain simpler numerical coefficients:

$$I = e^{2x}\left(\frac{2}{5}\sin x-\frac{1}{5}\cos x\right)+C$$

or, equivalently,

$$I = \frac{e^{2x}}{5}\bigl(2\sin x-\cos x\bigr)+C$$

where $$C$$ is the constant of integration.

Hence the integral is evaluated.

We have to evaluate the indefinite integral

$$I=\int \sin^{-1}\!\left(\dfrac{2x}{1+x^{2}}\right)\,\mathrm dx.$$

Step 1. A helpful substitution
Recall the trigonometric identity

$$\sin 2\theta = \dfrac{2\tan\theta}{1+\tan^{2}\theta}.$$

If we put

$$x = \tan\theta\qquad (\theta = \tan^{-1}x),$$

then

$$\frac{2x}{1+x^{2}} = \frac{2\tan\theta}{1+\tan^{2}\theta}=\sin 2\theta,$$ so that

$$\sin^{-1}\!\left(\dfrac{2x}{1+x^{2}}\right)=\sin^{-1}(\sin 2\theta)=2\theta,$$ where we use the principal value of $$\sin^{-1}$$ (for a Class 12 treatment we accept this equality; it is valid as long as $$2\theta\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$$, which certainly holds when $$x\ge 0$$ and can be adjusted for other values without affecting the result up to a constant).

We also need $$\mathrm dx$$ in terms of $$\theta$$:

$$x = \tan\theta \;\Rightarrow\; \mathrm dx = \sec^{2}\theta\,\mathrm d\theta = (1+\tan^{2}\theta)\,\mathrm d\theta = (1+x^{2})\,\mathrm d\theta.$$

Step 2. Transforming the integral

Substituting $$x=\tan\theta$$ gives

$$I = \int 2\theta\,\sec^{2}\theta\,\mathrm d\theta.$$

Step 3. Integration by parts
Choose

$$u = 2\theta, \quad \mathrm dv = \sec^{2}\theta\,\mathrm d\theta.$$

Then $$\mathrm du = 2\,\mathrm d\theta$$ and $$v = \tan\theta$$ (because $$\int \sec^{2}\theta\,\mathrm d\theta = \tan\theta$$).

Hence

$$I = u v - \int v\,\mathrm du = 2\theta\tan\theta - \int 2\tan\theta\,\mathrm d\theta.$$

But $$\int \tan\theta\,\mathrm d\theta = -\ln|\cos\theta|,$$ so

$$I = 2\theta\tan\theta - 2(-\ln|\cos\theta|) = 2\theta\tan\theta + 2\ln|\cos\theta| + C,$$ where $$C$$ is the constant of integration.

Step 4. Returning to the original variable
Replace $$\theta$$ and the trigonometric functions by $$x$$:

• $$\theta = \tan^{-1}x$$
• $$\tan\theta = x$$
• $$\cos\theta = \dfrac{1}{\sqrt{1+\tan^{2}\theta}} = \dfrac{1}{\sqrt{1+x^{2}}}$$

Therefore

$$2\ln|\cos\theta| = 2\ln\!\left|\dfrac{1}{\sqrt{1+x^{2}}}\right| = -\ln(1+x^{2}).$$

Substituting these in the expression for $$I$$, we get

$$I = 2x\tan^{-1}x - \ln(1+x^{2}) + C.$$

Step 5. (Optional) Differentiation check
Differentiate $$F(x)=2x\tan^{-1}x-\ln(1+x^{2})$$ to verify it yields the original integrand. A routine derivative calculation confirms that $$F'(x)=\sin^{-1}\!\bigl(\dfrac{2x}{1+x^{2}}\bigr).$$ Thus the result is correct.

Hence the required integral is

$$\boxed{\displaystyle \int \sin^{-1}\!\left(\dfrac{2x}{1+x^{2}}\right)\mathrm dx = 2x\tan^{-1}x - \ln(1+x^{2}) + C.}$$

We have to evaluate the indefinite integral $$\int x^2 e^{x^3}\,dx$$.

The exponential term is $$e^{x^3}$$ whose inner function is $$x^3$$. Since $$\dfrac{d}{dx}(x^3)=3x^2$$, the factor $$x^2$$ already present outside suggests the substitution method.

Substitution
Put $$t = x^3\,.$$ Then $$\dfrac{dt}{dx}=3x^2 \;\Longrightarrow\; dt = 3x^2\,dx \;\Longrightarrow\; x^2\,dx = \dfrac{dt}{3}.$$

Transform the integral
$$\int x^2 e^{x^3}\,dx = \int e^{t}\left(\dfrac{dt}{3}\right)=\dfrac{1}{3}\int e^{t}\,dt.$$

Integrate
The antiderivative of $$e^{t}$$ is itself, so
$$\dfrac{1}{3}\int e^{t}\,dt = \dfrac{1}{3}e^{t}+C.$$

Back-substitution
Replacing $$t$$ by $$x^3$$ gives
$$\int x^2 e^{x^3}\,dx = \dfrac{1}{3}e^{x^3}+C.$$

Hence the correct choice is (A).

The given integral is

$$I = \int e^x \sec x\,(1+\tan x)\,dx.$$

Notice that the integrand resembles the derivative of the product $$e^x\sec x.$$ To confirm this, differentiate $$e^x\sec x$$ using the product rule.

Step 1 : Differentiate.

$$\frac{d}{dx}\,(e^x\sec x)=e^x\sec x+e^x\,(\sec x\tan x).$$

Step 2 : Factor the common term.

$$=e^x\sec x\,\bigl(1+\tan x\bigr).$$

This derivative matches the integrand exactly:

$$e^x\sec x\,(1+\tan x).$$

Step 3 : Integrate.

Because the integrand is the derivative of $$e^x\sec x,$$ its integral is simply that function (plus the constant of integration):

$$\int e^x\sec x\,(1+\tan x)\,dx = e^x\sec x + C.$$

Hence the correct choice is (B).

Let $$I=\int \sqrt{x^{2}+2x+5}\,dx$$.

1. Complete the square
$$x^{2}+2x+5=(x+1)^{2}+4$$, so $$I=\int \sqrt{(x+1)^{2}+4}\,dx$$.

2. Trigonometric substitution
Put $$x+1=2\tan \theta\;\;(\,-\tfrac{\pi}{2}<\theta<\tfrac{\pi}{2}\,)$$.
Then $$dx=2\sec ^{2}\theta\,d\theta$$ and $$\sqrt{(x+1)^{2}+4}=\sqrt{4\tan ^{2}\theta+4}=2\sec \theta$$.

Hence

$$\begin{aligned} I &=\int \bigl(2\sec\theta\bigr)\bigl(2\sec ^{2}\theta\bigr)\,d\theta\\[2pt] &=4\int \sec ^{3}\theta\,d\theta. \end{aligned}$$

3. Integrate $$\sec ^{3}\theta$$
A standard result is
$$\int \sec ^{3}\theta\,d\theta=\tfrac12\sec \theta\tan \theta+\tfrac12\ln|\sec \theta+\tan \theta|+C.$$

Therefore

$$\begin{aligned} I &=4\Bigl[\tfrac12\sec \theta\tan \theta+\tfrac12\ln|\sec \theta+\tan \theta|\Bigr]+C\\[2pt] &=2\sec \theta\tan \theta+2\ln|\sec \theta+\tan \theta|+C. \end{aligned}$$

4. Re-substitute for $$\theta$$
From the substitution $$\tan \theta=\dfrac{x+1}{2}$$, so

$$\sec \theta=\sqrt{1+\tan ^{2}\theta}=\sqrt{1+\dfrac{(x+1)^{2}}{4}}=\dfrac{\sqrt{(x+1)^{2}+4}}{2}.$$

Hence

$$\begin{aligned} 2\sec \theta\tan \theta&=2\Bigl(\dfrac{\sqrt{(x+1)^{2}+4}}{2}\Bigr)\Bigl(\dfrac{x+1}{2}\Bigr)=\dfrac{(x+1)\sqrt{(x+1)^{2}+4}}{2},\\[4pt] 2\ln|\sec \theta+\tan \theta|&=2\ln\left|\dfrac{\sqrt{(x+1)^{2}+4}}{2}+\dfrac{x+1}{2}\right| =2\ln\left|\,x+1+\sqrt{(x+1)^{2}+4}\right|\;(\text{constant factor 2 absorbed}). \end{aligned}$$

5. Final result

$$I=\frac{(x+1)\sqrt{(x+1)^{2}+4}}{2}+2\ln\left|x+1+\sqrt{(x+1)^{2}+4}\right|+C.$$

Since $$(x+1)^{2}+4=x^{2}+2x+5$$, the answer may also be written as

$$\displaystyle \int\sqrt{x^{2}+2x+5}\,dx=\frac{(x+1)\sqrt{x^{2}+2x+5}}{2}+2\ln\left|x+1+\sqrt{x^{2}+2x+5}\right|+C.$$

Let

$$I = \int \sqrt{3-2x-x^2}\,dx.$$

1. Complete the square inside the radical

$$3-2x-x^2 = -\bigl(x^2+2x-3\bigr) = -\Bigl[(x+1)^2-4\Bigr] = 4-(x+1)^2.$$

Hence the integrand becomes $$\sqrt{4-(x+1)^2}.$$

2. Trigonometric substitution

Put $$x+1 = 2\sin\theta \quad\Longrightarrow\quad dx = 2\cos\theta\,d\theta,$$ where $$-\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}$$ so that $$\sin\theta$$ is one–to–one.

Then

$$\sqrt{4-(x+1)^2}=\sqrt{4-4\sin^2\theta}=\sqrt{4\cos^2\theta}=2\cos\theta.$$

Therefore

$$I = \int (2\cos\theta)\,(2\cos\theta\,d\theta)=\int 4\cos^2\theta\,d\theta.$$

3. Integrate in the \(\theta\)-variable

Use $$\cos^2\theta = \dfrac{1+\cos2\theta}{2}:$$

$$\begin{aligned} I &= \int 4\cos^2\theta\,d\theta \\ &= \int 4\cdot\frac{1+\cos2\theta}{2}\,d\theta \\ &= \int 2(1+\cos2\theta)\,d\theta \\ &= \int 2\,d\theta + \int 2\cos2\theta\,d\theta \\ &= 2\theta + \sin2\theta + C. \end{aligned}$$

4. Return to the variable \(x\)

• Because $$\sin\theta = \dfrac{x+1}{2},$$ we have $$\theta = \sin^{-1}\left(\dfrac{x+1}{2}\right).$$
• Compute $$\sin2\theta = 2\sin\theta\cos\theta.$$
  Here $$\cos\theta = \sqrt{1-\sin^2\theta}=\sqrt{1-\dfrac{(x+1)^2}{4}}=\dfrac{\sqrt{4-(x+1)^2}}{2}.$$
  Thus

$$\sin2\theta = 2\cdot\frac{x+1}{2}\cdot\frac{\sqrt{4-(x+1)^2}}{2}=\frac{(x+1)\sqrt{4-(x+1)^2}}{2}.$$

5. Final result

$$I = 2\sin^{-1}\left(\dfrac{x+1}{2}\right) + \frac{(x+1)}{2}\sqrt{4-(x+1)^2} + C.$$

Since $$4-(x+1)^2 = 3-2x-x^2,$$ the answer can also be written as

$$\boxed{\displaystyle \int \sqrt{3-2x-x^2}\,dx = 2\sin^{-1}\left(\dfrac{x+1}{2}\right) + \frac{(x+1)}{2}\sqrt{3-2x-x^2} + C.}$$

Let the required integral be

$$I = \int \sqrt{4 - x^{2}}\,dx$$

Step 1. Trigonometric substitution.
Put $$x = 2\sin\theta$$. Then

• $$dx = 2\cos\theta\,d\theta$$
• $$4 - x^{2} = 4 - 4\sin^{2}\theta = 4(1-\sin^{2}\theta) = 4\cos^{2}\theta$$
• $$\sqrt{4 - x^{2}} = \sqrt{4\cos^{2}\theta} = 2\cos\theta\quad(-\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}\;\Rightarrow\;\cos\theta \ge 0)$$

Therefore

$$I = \int \bigl(2\cos\theta\bigr)\bigl(2\cos\theta\,d\theta\bigr) = 4\int \cos^{2}\theta\,d\theta$$

Step 2. Integrate $$\cos^{2}\theta$$.
Use $$\cos^{2}\theta = \dfrac{1+\cos2\theta}{2}$$:

$$I = 4\int \dfrac{1+\cos2\theta}{2}\,d\theta = 2\int\bigl(1+\cos2\theta\bigr)\,d\theta$$

$$I = 2\left[ \theta + \tfrac{1}{2}\sin2\theta \right] + C = 2\theta + \sin2\theta + C$$

Step 3. Return to the variable $$x$$.

• Since $$x = 2\sin\theta\;\Rightarrow\;\theta = \sin^{-1}\!\left(\dfrac{x}{2}\right)$$
• $$\sin2\theta = 2\sin\theta\cos\theta = 2\left(\dfrac{x}{2}\right)\left(\dfrac{\sqrt{4-x^{2}}}{2}\right) = \dfrac{x\sqrt{4-x^{2}}}{2}$$

Substituting, we obtain

$$I = 2\sin^{-1}\!\left(\dfrac{x}{2}\right) + \dfrac{x\sqrt{4-x^{2}}}{2} + C$$

Thus,

$$\boxed{\displaystyle \int \sqrt{4 - x^{2}}\,dx = 2\sin^{-1}\!\left(\dfrac{x}{2}\right) + \dfrac{x\sqrt{4 - x^{2}}}{2} + C}$$

Step 1: Trigonometric substitution. The integrand $$\sqrt{1-4x^{2}}$$ is real for $$-\tfrac12\le x\le \tfrac12$$. Write the quadratic as $$1-4x^{2}=1-(2x)^{2}.$$ Put $$2x=\sin\theta$$ with $$-\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}$$ (so that $$\cos\theta\ge 0$$). Then $$x=\tfrac12\sin\theta$$ and $$dx=\tfrac12\cos\theta\,d\theta.$$

Step 2: Transform the integrand.

$$\sqrt{1-4x^{2}}=\sqrt{1-\sin^{2}\theta}=\cos\theta,$$

so

$$\int\sqrt{1-4x^{2}}\,dx=\int\cos\theta\cdot\tfrac12\cos\theta\,d\theta=\tfrac12\int\cos^{2}\theta\,d\theta.$$

Step 3: Integrate $$\cos^{2}\theta$$ using $$\cos^{2}\theta=\tfrac12(1+\cos 2\theta)$$:

$$\tfrac12\int\cos^{2}\theta\,d\theta=\tfrac14\int(1+\cos 2\theta)\,d\theta=\tfrac14\!\left(\theta+\tfrac12\sin 2\theta\right)+C.$$

Step 4: Return to $$x$$. With $$\sin\theta=2x$$ and $$\cos\theta=\sqrt{1-4x^{2}}$$, we have $$\theta=\sin^{-1}(2x)$$ and $$\sin 2\theta=2\sin\theta\cos\theta=4x\sqrt{1-4x^{2}}.$$ Therefore

$$\int\sqrt{1-4x^{2}}\,dx=\tfrac14\sin^{-1}(2x)+\tfrac18\cdot 4x\sqrt{1-4x^{2}}+C=\dfrac{x}{2}\sqrt{1-4x^{2}}+\tfrac14\sin^{-1}(2x)+C.$$

We have to evaluate the integral

$$\displaystyle\int \sqrt{x^{2}+4x+6}\,dx$$

1. Complete the square inside the root

$$x^{2}+4x+6=(x+2)^{2}+6-4=(x+2)^{2}+2$$

2. Substitute to simplify the integrand

Let $$t=x+2\;\Longrightarrow\;dt=dx$$. The integral becomes

$$\int\sqrt{t^{2}+2}\,dt$$

3. Recall the standard result

For any constant $$a>0$$,

$$\int\sqrt{t^{2}+a^{2}}\,dt=\frac{t}{2}\sqrt{t^{2}+a^{2}}+\frac{a^{2}}{2}\ln\left|t+\sqrt{t^{2}+a^{2}}\right|+C$$

Here $$a^{2}=2\;(\Rightarrow a=\sqrt2)$$. Applying the formula gives

$$\int\sqrt{t^{2}+2}\,dt=\frac{t}{2}\sqrt{t^{2}+2}+\frac{2}{2}\ln\left|t+\sqrt{t^{2}+2}\right|+C$$

$$=\tfrac12 t\sqrt{t^{2}+2}+\ln\left|t+\sqrt{t^{2}+2}\right|+C$$

4. Re-substitute $$t=x+2$$

$$\tfrac12 (x+2)\sqrt{(x+2)^{2}+2}+\ln\left|x+2+\sqrt{(x+2)^{2}+2}\right|+C$$

5. Replace the square-completed form by the original quadratic

Since $$(x+2)^{2}+2=x^{2}+4x+6$$, our final integrated expression is

$$\boxed{\,\tfrac12 (x+2)\sqrt{x^{2}+4x+6}+\ln\left|x+2+\sqrt{x^{2}+4x+6}\right|+C\,}$$

Integral to be evaluated

We wish to find $$\displaystyle\int \sqrt{x^{2}+4x+1}\,dx$$.

1. Complete the square

$$x^{2}+4x+1=(x+2)^{2}-3$$  because  $$x^{2}+4x+4-4+1=(x+2)^{2}-3$$.

2. A simpler variable

Put $$u=x+2\;\;(\Rightarrow\;du=dx)$$. Then

$$I=\int \sqrt{u^{2}-3}\,du$$.

3. Standard form – derivation once

Recall the standard result, obtained below once for clarity:

For a constant $$a>0$$,

$$\int \sqrt{u^{2}-a^{2}}\,du=\frac{u}{2}\sqrt{u^{2}-a^{2}}-\frac{a^{2}}{2}\ln\Bigl|u+\sqrt{u^{2}-a^{2}}\Bigr|+C.$$

Derivation (integration by parts)

• Write the integrand as $$\sqrt{u^{2}-a^{2}}=\bigl(u^{2}-a^{2}\bigr)^{1/2}$$.
• Choose $$f=\sqrt{u^{2}-a^{2}},\,g' =1\;(\Rightarrow g=u)$$ to apply integration by parts:
  $$\int f\,g'\,du=f\,g-\int g\,f'\,du.$$
• We need $$f'=\dfrac{d}{du}\sqrt{u^{2}-a^{2}}=\dfrac{u}{\sqrt{u^{2}-a^{2}}}.$$
• Therefore
  $$\int \sqrt{u^{2}-a^{2}}\,du=u\sqrt{u^{2}-a^{2}}-\int u\left(\dfrac{u}{\sqrt{u^{2}-a^{2}}}\right)du$$
  $$=u\sqrt{u^{2}-a^{2}}-\int \dfrac{u^{2}}{\sqrt{u^{2}-a^{2}}}\,du.$$
• Split the numerator: $$u^{2}=\bigl(u^{2}-a^{2}\bigr)+a^{2}$$. Hence
  $$\int \dfrac{u^{2}}{\sqrt{u^{2}-a^{2}}}\,du=\int \sqrt{u^{2}-a^{2}}\,du+a^{2}\int \dfrac{du}{\sqrt{u^{2}-a^{2}}}.$$
• Denote the original integral by $$I$$; then we have
  $$I=u\sqrt{u^{2}-a^{2}}-\bigl(I+a^{2}\int \dfrac{du}{\sqrt{u^{2}-a^{2}}}\bigr).$$
• Re-arranging, $$2I=u\sqrt{u^{2}-a^{2}}-a^{2}\int \dfrac{du}{\sqrt{u^{2}-a^{2}}}.$$
• The remaining integral is elementary: $$\int \dfrac{du}{\sqrt{u^{2}-a^{2}}}=\ln\Bigl|u+\sqrt{u^{2}-a^{2}}\Bigr|+C.$$
• Hence $$2I=u\sqrt{u^{2}-a^{2}}-a^{2}\ln\Bigl|u+\sqrt{u^{2}-a^{2}}\Bigr|,$$ giving the claimed formula after dividing by 2.

4. Apply with $$a=\sqrt{3}$$

Using the formula for $$a^{2}=3$$, we get

$$I=\frac{u}{2}\sqrt{u^{2}-3}-\frac{3}{2}\ln\Bigl|u+\sqrt{u^{2}-3}\Bigr|+C.$$

5. Return to $$x$$

Recall $$u=x+2$$ and $$\sqrt{u^{2}-3}=\sqrt{(x+2)^{2}-3}=\sqrt{x^{2}+4x+1}$$. Therefore

$$\boxed{\;\displaystyle\int \sqrt{x^{2}+4x+1}\,dx=\frac{x+2}{2}\,\sqrt{x^{2}+4x+1}-\frac{3}{2}\,\ln\Bigl|(x+2)+\sqrt{x^{2}+4x+1}\Bigr|+C\;}$$.

Integrate

$$I = \int \sqrt{1-4x-x^2}\,dx$$

1. Complete the square inside the radical

$$1-4x-x^2 = -(x^2+4x-1)$$

Write the quadratic in completed-square form:

$$x^2+4x-1 = (x+2)^2-4-1=(x+2)^2-5$$

Therefore

$$1-4x-x^2 = 5-(x+2)^2$$

2. Substitute to centre the square

Let $$u = x+2 \;\Rightarrow\; du = dx$$. Then

$$I = \int \sqrt{5-u^2}\,du$$

3. Use the standard integral

The well-known formula

$$\int \sqrt{a^2-u^2}\,du = \frac{u}{2}\sqrt{a^2-u^2}+\frac{a^2}{2}\sin^{-1}\!\left(\frac{u}{a}\right)+C$$

applies with $$a^2=5\;(\,a=\sqrt5\,)$$. Hence

$$I = \frac{u}{2}\sqrt{5-u^2}+\frac{5}{2}\sin^{-1}\!\left(\frac{u}{\sqrt5}\right)+C$$

4. Replace $$u$$ by $$x$$

Recall $$u=x+2$$:

$$I = \frac{x+2}{2}\,\sqrt{5-(x+2)^2}+\frac{5}{2}\sin^{-1}\!\left(\frac{x+2}{\sqrt5}\right)+C$$

5. Express the radical as in the original integrand (optional)

Because $$\sqrt{5-(x+2)^2}=\sqrt{1-4x-x^2}$$, the result can also be written as

$$I = \frac{x+2}{2}\,\sqrt{1-4x-x^2}+\frac{5}{2}\sin^{-1}\!\left(\frac{x+2}{\sqrt5}\right)+C$$

This is the required antiderivative.

Integral to be evaluated: $$\displaystyle \int \sqrt{x^{2}+4x-5}\,dx$$.

Step 1: Complete the square in the quadratic expression.

Write $$x^{2}+4x-5=(x^{2}+4x+4)-4-5=(x+2)^{2}-9.$$

Hence $$\sqrt{x^{2}+4x-5}=\sqrt{(x+2)^{2}-3^{2}}.$$

Step 2: Substitute $$u=x+2$$.

Then $$du=dx$$, and the integral becomes $$\displaystyle I=\int\sqrt{u^{2}-9}\,du.$$

Step 3: Integrate $$\int\sqrt{u^{2}-a^{2}}\,du$$ for the case $$a=3$$.

A convenient trigonometric substitution is $$u=3\sec\theta$$ (choose the sign of $$\theta$$ so that $$|u|\ge 3$$). Then

• $$du=3\sec\theta\tan\theta\,d\theta,$$
• $$\sqrt{u^{2}-9}=\sqrt{9\sec^{2}\theta-9}=3\tan\theta.$$

Therefore

$$I=\int \bigl(3\tan\theta\bigr)\bigl(3\sec\theta\tan\theta\bigr)\,d\theta=9\int\sec\theta\,\tan^{2}\theta\,d\theta.$$

Using $$\tan^{2}\theta=\sec^{2}\theta-1$$,

$$I=9\int\bigl(\sec^{3}\theta-\sec\theta\bigr)\,d\theta=9\Bigl(\int\sec^{3}\theta\,d\theta-\int\sec\theta\,d\theta\Bigr).$$

The standard results

• $$\int\sec\theta\,d\theta=\ln\bigl|\sec\theta+\tan\theta\bigr|+C_{1},$$
• $$\int\sec^{3}\theta\,d\theta=\tfrac12\bigl(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|\bigr)+C_{2}$$

give

$$\begin{aligned} I &=9\left[\tfrac12\bigl(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|\bigr)-\ln|\sec\theta+\tan\theta|\right]+C \\ &=\tfrac{9}{2}\sec\theta\tan\theta-\tfrac{9}{2}\ln|\sec\theta+\tan\theta|+C. \end{aligned}$$

Step 4: Return to the variable $$u$$.

Because $$u=3\sec\theta$$, we have $$\sec\theta=\dfrac{u}{3}$$ and $$\tan\theta=\sqrt{\sec^{2}\theta-1}=\dfrac{\sqrt{u^{2}-9}}{3}$$. Thus

$$\sec\theta\tan\theta=\dfrac{u}{3}\cdot\dfrac{\sqrt{u^{2}-9}}{3}=\dfrac{u\sqrt{u^{2}-9}}{9}.$$

Substituting in the expression for $$I$$,

$$I=\frac{9}{2}\cdot\frac{u\sqrt{u^{2}-9}}{9}-\frac{9}{2}\ln\bigl|\,\tfrac{u}{3}+\tfrac{\sqrt{u^{2}-9}}{3}\bigr|+C =\frac{u}{2}\sqrt{u^{2}-9}-\frac{9}{2}\ln|u+\sqrt{u^{2}-9}|+C'.$$

(The constant $$-\tfrac{9}{2}\ln 3$$ has been absorbed into the overall constant of integration $$C'$$.)

Step 5: Substitute back $$u=x+2$$.

$$\int\sqrt{x^{2}+4x-5}\,dx=\frac{x+2}{2}\sqrt{x^{2}+4x-5}-\frac{9}{2}\ln\left|x+2+\sqrt{x^{2}+4x-5}\right|+C.$$

Domain note. The integrand is real when $$x^{2}+4x-5\ge 0$$, i.e. when $$x\le -5$$ or $$x\ge 1$$. The antiderivative obtained above is valid on each of these intervals.

Integral to be evaluated :  $$I = \displaystyle\int \sqrt{1+3x-x^2}\,dx$$

1. Complete the square in the radicand

$$1+3x-x^2 \;=\; -(x^2-3x-1)$$
Rewrite the quadratic inside the brackets:

$$x^2-3x-1 \,=\, \bigl(x-\tfrac{3}{2}\bigr)^2 -\tfrac{9}{4}-1$$

Therefore

$$-(x^2-3x-1)= -\Bigl( \bigl(x-\tfrac{3}{2}\bigr)^2-\tfrac{13}{4}\Bigr)=\tfrac{13}{4}-\bigl(x-\tfrac{3}{2}\bigr)^2$$

Hence the integrand becomes

$$\sqrt{1+3x-x^2}=\sqrt{\tfrac{13}{4}-\bigl(x-\tfrac{3}{2}\bigr)^2}$$

2. Introduce a simple variable

Put $$u = x-\tfrac{3}{2}\;\;\Rightarrow\;\;du = dx$$ and denote $$a^2 = \tfrac{13}{4}\;\;(a=\tfrac{\sqrt{13}}{2}).$$

Then

$$I = \int \sqrt{a^2-u^2}\,du$$

3. Use the standard integral

A standard result (proved by substituting $$u=a\sin\theta$$) is

$$\int \sqrt{a^2-u^2}\,du = \tfrac{u}{2}\sqrt{a^2-u^2}+\tfrac{a^2}{2}\sin^{-1}\!\bigl(\tfrac{u}{a}\bigr)+C.$$

Applying it with the present $$u$$ and $$a$$:

$$I = \tfrac{u}{2}\sqrt{a^2-u^2}+\tfrac{a^2}{2}\sin^{-1}\!\bigl(\tfrac{u}{a}\bigr)+C.$$

4. Substitute back

Replace $$u$$ and $$a$$ by $$x-\tfrac{3}{2}$$ and $$\tfrac{\sqrt{13}}{2}$$ respectively.

First term:

$$\tfrac{u}{2}\sqrt{a^2-u^2}=\tfrac{x-\tfrac{3}{2}}{2}\,\sqrt{\tfrac{13}{4}-\bigl(x-\tfrac{3}{2}\bigr)^2}=\tfrac{x-\tfrac{3}{2}}{2}\sqrt{1+3x-x^2}.$$

Second term:

$$\tfrac{a^2}{2}\sin^{-1}\!\bigl(\tfrac{u}{a}\bigr)=\tfrac{\tfrac{13}{4}}{2}\sin^{-1}\!\Bigl(\tfrac{2(x-\tfrac{3}{2})}{\sqrt{13}}\Bigr)=\tfrac{13}{8}\sin^{-1}\!\Bigl(\tfrac{2x-3}{\sqrt{13}}\Bigr).$$

5. Final result

$$\boxed{\displaystyle\int \sqrt{1+3x-x^2}\,dx = \frac{x-\tfrac{3}{2}}{2}\,\sqrt{1+3x-x^2}\; +\; \frac{13}{8}\,\sin^{-1}\!\Bigl(\frac{2x-3}{\sqrt{13}}\Bigr)+C}$$

We have to evaluate

$$I = \int \sqrt{x^{2}+3x}\,dx.$$

1. Complete the square.

$$x^{2}+3x = \Bigl(x+\tfrac{3}{2}\Bigr)^{2}-\Bigl(\tfrac{3}{2}\Bigr)^{2}.$$

2. Substitute $$u = x+\tfrac{3}{2}$$ so $$du = dx$$:

$$I = \int \sqrt{u^{2}-\Bigl(\tfrac{3}{2}\Bigr)^{2}}\,du.$$

3. Apply the standard formula $$\int \sqrt{u^{2}-a^{2}}\,du = \dfrac{u}{2}\sqrt{u^{2}-a^{2}} - \dfrac{a^{2}}{2}\ln\!\left|u+\sqrt{u^{2}-a^{2}}\right| + C$$ with $$a = \tfrac32$$ (so $$a^{2}=\tfrac94$$):

$$I = \dfrac{u}{2}\sqrt{u^{2}-\tfrac{9}{4}} - \dfrac{9}{8}\ln\!\left|u+\sqrt{u^{2}-\tfrac{9}{4}}\right| + C.$$

4. Return to $$x$$. Note that $$\sqrt{u^{2}-\tfrac94}=\sqrt{(x+\tfrac32)^{2}-\tfrac94}=\sqrt{x^{2}+3x}.$$ Hence

$$I = \dfrac{1}{2}\Bigl(x+\tfrac{3}{2}\Bigr)\sqrt{x^{2}+3x} - \dfrac{9}{8}\ln\!\left|x+\tfrac{3}{2}+\sqrt{x^{2}+3x}\right| + C.$$

We have to evaluate the integral

$$I = \int \sqrt{1+\dfrac{x^{2}}{9}}\,dx$$

Step 1 : Reduce the integrand.

$$\sqrt{1+\dfrac{x^{2}}{9}} = \sqrt{\dfrac{9+x^{2}}{9}} = \dfrac{1}{3}\,\sqrt{x^{2}+9}$$

Hence

$$I = \dfrac13\,\int \sqrt{x^{2}+9}\,dx$$

Step 2 : Recall the standard integral.

For any constant $$a>0$$,

$$\int \sqrt{x^{2}+a^{2}}\,dx = \dfrac{x}{2}\sqrt{x^{2}+a^{2}} + \dfrac{a^{2}}{2}\ln\left|x+\sqrt{x^{2}+a^{2}}\right| + C$$

This can be proved quickly with the substitution $$x=a\tan\theta$$ (a proof is shown in Step 3). Here $$a=3$$, so

$$\int \sqrt{x^{2}+9}\,dx = \dfrac{x}{2}\sqrt{x^{2}+9} + \dfrac{9}{2}\ln\left|x+\sqrt{x^{2}+9}\right| + C$$

Step 3 : (Optional) Derivation of the standard result.

Put $$x=3\tan\theta \;\;(\theta\in\left(-\dfrac\pi2,\dfrac\pi2\right))$$. Then $$dx = 3\sec^{2}\theta\,d\theta$$ and

$$\sqrt{x^{2}+9}=\sqrt{9\tan^{2}\theta+9}=3\sec\theta$$

Therefore

$$\int\sqrt{x^{2}+9}\,dx = \int 3\sec\theta\,(3\sec^{2}\theta)\,d\theta = 9\int\sec^{3}\theta\,d\theta$$

Using the well-known result $$\int\sec^{3}\theta\,d\theta = \dfrac12(\sec\theta\tan\theta)+\dfrac12\ln|\sec\theta+\tan\theta|+C$$, we obtain

$$9\Bigl[\dfrac12(\sec\theta\tan\theta)+\dfrac12\ln|\sec\theta+\tan\theta|\Bigr]+C$$

Now \;$$\tan\theta=\dfrac{x}{3}, \;\sec\theta=\dfrac{\sqrt{x^{2}+9}}{3}$$, so

$$\sec\theta\tan\theta = \dfrac{x\sqrt{x^{2}+9}}{9}, \qquad \ln|\sec\theta+\tan\theta| = \ln\left|\dfrac{\sqrt{x^{2}+9}+x}{3}\right|$$

Putting these back and absorbing the constant $$\ln3$$ into $$C$$ produces exactly the standard result quoted above.

Step 4 : Substitute the standard result into $$I$$.

$$I = \dfrac13\Bigl[\dfrac{x}{2}\sqrt{x^{2}+9} + \dfrac{9}{2}\ln\left|x+\sqrt{x^{2}+9}\right|\Bigr] + C$$

Simplify the numerical factors:

$$\dfrac13\cdot\dfrac{x}{2}=\dfrac{x}{6},\qquad \dfrac13\cdot\dfrac{9}{2}=\dfrac32$$

Thus

$$I = \dfrac{x\sqrt{x^{2}+9}}{6}+\dfrac{3}{2}\ln\left|x+\sqrt{x^{2}+9}\right|+C$$

Final Result.

$$\boxed{\displaystyle \int\sqrt{1+\dfrac{x^{2}}{9}}\,dx = \dfrac{x\sqrt{x^{2}+9}}{6}+\dfrac{3}{2}\ln\left|x+\sqrt{x^{2}+9}\right|+C}$$

We have to evaluate the integral $$\int \sqrt{1+x^{2}}\,dx$$ and then identify which of the four given expressions is its antiderivative.

Step 1 – Choose a suitable substitution
Because the integrand contains $$1+x^{2}$$ under a square-root, the standard hyperbolic substitution $$x=\sinh t$$ is convenient.

Put $$x=\sinh t\;\;(t\in\mathbb R).$$ Then $$dx=\cosh t\,dt$$ and

$$1+x^{2}=1+\sinh ^2 t=\cosh ^2 t\;\;\bigl(\because\;1+\sinh ^2t=\cosh ^2t\bigr).$$

Hence

$$\sqrt{1+x^{2}}=\cosh t.$$

Step 2 – Transform the integral

\[ \int \sqrt{1+x^{2}}\,dx \;=\;\int \cosh t\,(\cosh t\,dt) \;=\;\int \cosh ^2 t\,dt. \]

Step 3 – Integrate $$\cosh ^2 t$$

Recall the double-angle identity $$\cosh 2t=2\cosh ^2t-1\;\Longrightarrow\;\cosh ^2 t=\dfrac{\cosh 2t+1}{2}.$$

Therefore

\[ \int \cosh ^2 t\,dt \;=\;\int\frac{\cosh 2t+1}{2}\,dt \;=\;\frac12\int\cosh 2t\,dt+\frac12\int dt. \]

The two integrals are elementary:

• $$\int\cosh 2t\,dt=\tfrac12\sinh 2t$$ (because the derivative of $$\sinh 2t$$ is $$2\cosh 2t$$),
• $$\int dt=t.$$

Putting these back,

\[ \int \cosh ^2 t\,dt =\frac12\Bigl(\tfrac12\sinh 2t\Bigr)+\frac12 t =\frac14\sinh 2t+\frac12 t+C. \]

Step 4 – Return to the variable $$x$$

We already have $$x=\sinh t\;\Rightarrow\;t=\sinh ^{-1}x.$$ Also, $$\sinh 2t=2\sinh t\cosh t=2x\sqrt{1+x^{2}}.$$ Substitute these:

$$\frac14\sinh 2t=\frac14(2x\sqrt{1+x^{2}})=\frac{x\sqrt{1+x^{2}}}{2}.$$

Further, the inverse hyperbolic relation gives

$$t=\sinh ^{-1}x=\ln\left|x+\sqrt{1+x^{2}}\right|.$$

Therefore

\[ \int \sqrt{1+x^{2}}\,dx =\frac{x}{2}\sqrt{1+x^{2}}+\frac12\ln\left|x+\sqrt{1+x^{2}}\right|+C. \]

Step 5 – Match with the options

The expression obtained is exactly option (A):
$$\displaystyle\tfrac{x}{2}\sqrt{1+x^{2}}+\tfrac12\log\left|x+\sqrt{1+x^{2}}\right|+C.$$

Hence the correct choice is (A).

We have to evaluate

$$I=\int \sqrt{x^{2}-8x+7}\,dx.$$

1. Complete the square inside the root

$$x^{2}-8x+7=(x^{2}-8x+16)-16+7=(x-4)^{2}-9.$$

2. Substitute to simplify

Put $$u=x-4\;\;(\;\Rightarrow\;x=u+4,\;dx=du).$$
Then

$$I=\int\sqrt{u^{2}-9}\,du.$$

3. Use the standard integral

For a positive constant $$a,$$

$$\int\sqrt{u^{2}-a^{2}}\,du=\frac{u}{2}\sqrt{u^{2}-a^{2}}-\frac{a^{2}}{2}\,\ln\left|u+\sqrt{u^{2}-a^{2}}\right|+C.$$

Here $$a=3$$ (because $$a^{2}=9$$). Applying the formula:

$$I=\frac{u}{2}\sqrt{u^{2}-9}-\frac{9}{2}\ln\left|u+\sqrt{u^{2}-9}\right|+C.$$

4. Return to the original variable

Replace $$u=x-4$$ and $$\sqrt{u^{2}-9}=\sqrt{(x-4)^{2}-9}=\sqrt{x^{2}-8x+7}.$$ Thus

$$I=\frac{1}{2}(x-4)\sqrt{x^{2}-8x+7}-\frac{9}{2}\ln\left|x-4+\sqrt{x^{2}-8x+7}\right|+C.$$

5. Match with the given options

This expression is exactly the one given in option (D).

We have to evaluate the definite integral

$$\int_{-1}^{1} \bigl(x+1\bigr)\,dx$$

Step 1: Find an antiderivative.

Integrate term by term:

$$\int x\,dx = \frac{x^{2}}{2}, \qquad \int 1\,dx = x.$$

Hence one antiderivative of $$x+1$$ is

$$F(x)=\frac{x^{2}}{2}+x.$$

Step 2: Apply the Fundamental Theorem of Calculus.

$$\int_{-1}^{1} (x+1)\,dx = F(1) - F(-1).$$

Compute each value:

$$F(1)=\frac{1^{2}}{2}+1 = \frac12 + 1 = \frac32,$$

$$F(-1)=\frac{(-1)^{2}}{2}+(-1) = \frac12 - 1 = -\frac12.$$

Now subtract:

$$F(1)-F(-1)=\frac32 - \bigl(-\tfrac12\bigr)=\frac32+\frac12=\frac{4}{2}=2.$$

Therefore

$$\int_{-1}^{1} (x+1)\,dx = 2.$$

We need to evaluate the definite integral

$$\displaystyle\int_{2}^{3}\frac{1}{x}\,dx$$

Step 1: Find an antiderivative.
Since $$\dfrac{d}{dx}(\ln|x|)=\dfrac{1}{x},$$ an antiderivative of $$\dfrac{1}{x}$$ is $$F(x)=\ln|x|$$.

Step 2: Apply the Fundamental Theorem of Calculus.
$$\displaystyle\int_{2}^{3}\frac{1}{x}\,dx = F(3) - F(2).$$

Step 3: Substitute the antiderivative values.
$$F(3)-F(2)=\ln|3| - \ln|2|.$$
Because both limits are positive, the absolute value signs are unnecessary:

$$\ln 3 - \ln 2.$$

Step 4: Simplify using logarithm properties.
$$\ln 3 - \ln 2 = \ln\left(\dfrac{3}{2}\right).$$

Hence

$$\displaystyle\int_{2}^{3}\frac{1}{x}\,dx = \ln\left(\dfrac{3}{2}\right).$$

We have to compute the definite integral

$$\int_{1}^{2}\,(4x^{3}-5x^{2}+6x+9)\,dx$$

Step 1 : Find an antiderivative.

Integrate each term separately:

• $$\int 4x^{3}\,dx = 4 \cdot \dfrac{x^{4}}{4}=x^{4}$$
• $$\int (-5x^{2})\,dx = -5\cdot \dfrac{x^{3}}{3}= -\dfrac{5x^{3}}{3}$$
• $$\int 6x\,dx = 6\cdot \dfrac{x^{2}}{2}=3x^{2}$$
• $$\int 9\,dx = 9x$$

Thus an antiderivative is

$$F(x)=x^{4}-\dfrac{5x^{3}}{3}+3x^{2}+9x$$

Step 2 : Evaluate at the limits.

At $$x=2$$:

$$F(2)=2^{4}-\dfrac{5(2)^{3}}{3}+3(2)^{2}+9(2)=16-\dfrac{40}{3}+12+18$$

Combine the numbers:

$$16+12+18=46 \;\Rightarrow\; F(2)=46-\dfrac{40}{3}=\dfrac{138}{3}-\dfrac{40}{3}=\dfrac{98}{3}$$

At $$x=1$$:

$$F(1)=1^{4}-\dfrac{5(1)^{3}}{3}+3(1)^{2}+9(1)=1-\dfrac{5}{3}+3+9$$

Combine the numbers:

$$1+3+9=13 \;\Rightarrow\; F(1)=13-\dfrac{5}{3}=\dfrac{39}{3}-\dfrac{5}{3}=\dfrac{34}{3}$$

Step 3 : Compute the definite integral.

$$\int_{1}^{2}(4x^{3}-5x^{2}+6x+9)\,dx = F(2)-F(1)=\dfrac{98}{3}-\dfrac{34}{3}=\dfrac{64}{3}$$

Hence, the value of the integral is $$\dfrac{64}{3}$$.

We have to evaluate the definite integral

$$\displaystyle\int_{0}^{\pi/4}\sin 2x\,dx$$

Step 1 : Find an antiderivative

Recall that for any constant $$a$$, $$\int \sin ax\,dx=-\dfrac{\cos ax}{a}+C$$, because $$\dfrac{d}{dx}\bigl(\cos ax\bigr)=-a\sin ax$$.

Taking $$a=2$$, an antiderivative of $$\sin 2x$$ is therefore

$$-\dfrac{\cos 2x}{2}$$.

Step 2 : Apply the Fundamental Theorem of Calculus

$$\displaystyle\int_{0}^{\pi/4}\sin 2x\,dx\;=\;\Bigl[-\dfrac{\cos 2x}{2}\Bigr]_{0}^{\pi/4}$$

Evaluate at the bounds:

Upper limit $$x=\dfrac{\pi}{4}$$:

$$-\dfrac{\cos 2\left(\dfrac{\pi}{4}\right)}{2}= -\dfrac{\cos \dfrac{\pi}{2}}{2}= -\dfrac{0}{2}=0.$$

Lower limit $$x=0$$:

$$-\dfrac{\cos 2\cdot 0}{2}= -\dfrac{\cos 0}{2}= -\dfrac{1}{2}.$$

Subtract the value at the lower limit from that at the upper limit:

$$0-\bigl(-\dfrac{1}{2}\bigr)=\dfrac{1}{2}.$$

Hence the value of the definite integral is $$\dfrac{1}{2}$$.

Given integral:

$$I = \displaystyle\int_{0}^{\pi/2} \cos 2x\,dx$$

Step 1: Find an antiderivative of $$\cos 2x$$.

Recall the standard result $$\int \cos kx\,dx = \dfrac{1}{k}\sin kx + C$$, where $$k$$ is a non–zero constant.

Here $$k = 2$$, so

$$\int \cos 2x\,dx = \dfrac{1}{2}\sin 2x + C.$$

Step 2: Apply the limits $$x = 0$$ to $$x = \pi/2$$ using the Fundamental Theorem of Calculus.

$$I = \Bigl[\dfrac{1}{2}\sin 2x\Bigr]_{0}^{\pi/2}$$

$$\;\;\; = \dfrac{1}{2}\bigl(\sin(2\,(\pi/2))\; - \; \sin(2 \times 0)\bigr)$$

Compute the sine values:

• $$\sin(2\,(\pi/2)) = \sin \pi = 0$$
• $$\sin(0) = 0$$

Therefore,

$$I = \dfrac{1}{2}\,(0 - 0) = 0.$$

Conclusion: The value of the definite integral is $$0$$.

We have to evaluate the definite integral

$$\int_{4}^{5} e^{x}\,dx$$

Step 1 – Find an antiderivative of the integrand.

The integrand is the exponential function $$e^{x}$$. Its derivative is itself; hence an antiderivative is also $$e^{x}$$, because

$$\frac{d}{dx}\,(e^{x}) = e^{x}.$$

Step 2 – Apply the Fundamental Theorem of Calculus.

If $$F(x)$$ is an antiderivative of $$f(x)$$, then

$$\int_{a}^{b} f(x)\,dx = F(b) - F(a).$$

Here, $$F(x)=e^{x}$$, $$a = 4$$ and $$b = 5$$, so

$$\int_{4}^{5} e^{x}\,dx = e^{5} - e^{4}.$$

Step 3 – Present the result in a simplified form.

We can leave the answer as $$e^{5} - e^{4}$$, or factor out $$e^{4}$$:

$$e^{5} - e^{4} = e^{4}(e - 1).$$

Optional numerical value: using $$e \approx 2.71828$$,

$$e^{5} - e^{4} \approx 148.41316 - 54.59815 \approx 93.81501.$$

However, the exact value is preferred in calculus problems.

Therefore,

$$\boxed{\displaystyle\int_{4}^{5} e^{x}\,dx = e^{5} - e^{4} = e^{4}(e - 1).}$$

We have to evaluate the definite integral

$$I = \int_{0}^{\pi/4} \tan x\,dx$$

Step 1. Find an antiderivative of $$\tan x$$.

Recall that $$\tan x = \dfrac{\sin x}{\cos x}$$ and that the derivative of $$\cos x$$ is $$-\sin x$$. Therefore:

$$\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx$$

Put $$u = \cos x \;\Rightarrow\; du = -\sin x\,dx \;\Rightarrow\; -du = \sin x\,dx$$.

Then

$$\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx = -\int \frac{du}{u} = -\ln|u| + C = -\ln|\cos x| + C.$$

Thus an antiderivative is $$-\ln|\cos x|$$ (equivalently $$\ln|\sec x|$$).

Step 2. Apply the limits $$0$$ and $$\dfrac{\pi}{4}$$.

$$I = \left[-\ln|\cos x|\right]_{0}^{\pi/4} = \bigl(-\ln|\cos(\pi/4)|\bigr)\; - \;\bigl(-\ln|\cos 0|\bigr).$$

Upper limit: $$\cos\left(\dfrac{\pi}{4}\right) = \dfrac{\sqrt2}{2} = \dfrac1{\sqrt2}.$$ Hence

$$-\ln\left(\dfrac{\sqrt2}{2}\right)= -\ln\left(\dfrac1{\sqrt2}\right)= -\left(-\ln\sqrt2\right)= \ln\sqrt2.$$

Because $$\ln\sqrt2 = \ln\left(2^{1/2}\right)= \dfrac12\ln2$$, the upper-limit value is $$\dfrac12\ln2.$$

Lower limit: $$\cos0 = 1 \;\Rightarrow\; -\ln 1 = 0.$$

Step 3. Subtract.

$$I = \dfrac12\ln2 - 0 = \dfrac12\ln2.$$

Therefore,

$$\boxed{\displaystyle\int_{0}^{\pi/4} \tan x\,dx = \dfrac12\ln 2}.$$

We are required to evaluate the definite integral

$$I= \int_{\pi/6}^{\pi/4}\csc x\,dx.$$

1. Antiderivative of $$\csc x$$

Multiply and divide the integrand by $$(\csc x+\cot x)$$:

$$\int \csc x\,dx = \int \csc x\,\frac{\csc x+\cot x}{\csc x+\cot x}\,dx = \int \frac{\csc^2 x+\csc x\cot x}{\csc x+\cot x}\,dx.$$

Let $$u = \csc x + \cot x.$$ Then $$du = -\big(\csc^2 x+\csc x\cot x\big)\,dx,$$ or equivalently $$-(du) = (\csc^2 x+\csc x\cot x)\,dx.$$

Therefore

$$\int \csc x\,dx = -\int \frac{du}{u}= -\ln|u|+C.$$

Substituting back,

$$\int \csc x\,dx = -\ln|\csc x+\cot x|+C = \ln|\tan (x/2)|+C.$$

(Both forms differ only by an additive constant.)

2. Apply the limits

$$I = \left[\ln\big|\tan (x/2)\big|\right]_{x=\pi/6}^{x=\pi/4} = \ln\left|\tan\frac{\pi}{8}\right| - \ln\left|\tan\frac{\pi}{12}\right|.$$

3. Exact values of the tangents

• $$\tan \frac{\pi}{8}=\sqrt{2}-1$$ (half–angle of $$45^{\circ}$$).
• $$\tan \frac{\pi}{12}=2-\sqrt{3}$$ ($$45^{\circ}-30^{\circ}$$ identity).

Hence

$$I = \ln\left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right).$$

Since $$(2-\sqrt{3})(2+\sqrt{3})=1,$$ rationalising the denominator gives

$$I = \ln\big[(\sqrt{2}-1)(2+\sqrt{3})\big].$$

4. Result

$$\boxed{\,\ln\big((\sqrt{2}-1)(2+\sqrt{3})\big)\,}$$

We have to evaluate the definite integral $$\displaystyle\int_0^1 \dfrac{dx}{\sqrt{1 - x^2}}$$.

Step 1 – Substitution:
Put $$x = \sin \theta\;,$$ where $$\theta\in[0,\tfrac{\pi}{2}]$$ because the interval $$x\in[0,1]$$ corresponds to the first quadrant.

Then
$$dx = \cos \theta\,d\theta$$ and $$\sqrt{1 - x^2}=\sqrt{1-\sin^2\theta}=\sqrt{\cos^2\theta}=\cos\theta\;.$$

Step 2 – Change of limits:
When $$x = 0$$, $$\sin\theta = 0 \Rightarrow \theta = 0.$$
When $$x = 1$$, $$\sin\theta = 1 \Rightarrow \theta = \dfrac{\pi}{2}.$$

Step 3 – Transform the integral:
$$\int_0^1 \dfrac{dx}{\sqrt{1 - x^2}} = \int_{0}^{\pi/2} \dfrac{\cos\theta\,d\theta}{\cos\theta} = \int_{0}^{\pi/2} d\theta.$$

Step 4 – Integrate and apply limits:
$$\int_{0}^{\pi/2} d\theta = \Big[\theta\Big]_{0}^{\pi/2}= \dfrac{\pi}{2}-0 = \dfrac{\pi}{2}.$$

Therefore,
$$\int_0^1 \dfrac{dx}{\sqrt{1 - x^2}} = \dfrac{\pi}{2}.$$

Step 1 – Recall a standard antiderivative
We know that the derivative of $$\tan^{-1}x$$ is $$\dfrac{1}{1+x^{2}}$$, so
$$\int \dfrac{dx}{1+x^{2}} = \tan^{-1}x + C.$$

Step 2 – Apply the limits 0 and 1
Using the Fundamental Theorem of Calculus,

Step 3 – Evaluate at the bounds

• Upper limit: $$\tan^{-1}1 = \dfrac{\pi}{4}$$ because $$\tan\dfrac{\pi}{4}=1$$.
• Lower limit: $$\tan^{-1}0 = 0$$ because $$\tan 0 = 0$$.

Step 4 – Subtract
$$\bigl[\tan^{-1}x\bigr]_{0}^{1} = \dfrac{\pi}{4} - 0 = \dfrac{\pi}{4}.$$

Result: $$\displaystyle \int_{0}^{1} \dfrac{dx}{1+x^{2}} = \dfrac{\pi}{4}.$$

The integrand is continuous on $$[2,3]$$ because $$x^2-1\ne 0$$ there, so the integral exists.

Write the integrand in partial fractions:

$$\frac{1}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}.$$

Multiplying by $$(x-1)(x+1)$$: $$1=A(x+1)+B(x-1).$$ Substituting roots:

• $$x=1$$:  $$1=2A \Rightarrow A=\tfrac12.$$
• $$x=-1$$:  $$1=-2B \Rightarrow B=-\tfrac12.$$

Thus

$$\frac{1}{x^2-1}=\tfrac12\!\left(\frac{1}{x-1}-\frac{1}{x+1}\right).$$

Integrating,

$$\int\frac{dx}{x^2-1}=\tfrac12\bigl(\ln|x-1|-\ln|x+1|\bigr)+C = \tfrac12\ln\!\left|\frac{x-1}{x+1}\right|+C.$$

Now evaluate from $$x=2$$ to $$x=3$$:

$$I = \tfrac12\ln\!\left|\frac{x-1}{x+1}\right|\bigg|_{2}^{3} = \tfrac12\!\left(\ln\tfrac12 - \ln\tfrac13\right) = \tfrac12\ln\tfrac32.$$

Therefore

$$\int_2^3\frac{dx}{x^2-1}=\tfrac12\ln\!\left(\tfrac32\right).$$

We have to evaluate the definite integral

$$I=\displaystyle\int_{0}^{\pi/2}\cos^{2}x\,dx.$$

Step 1: Use a power-reducing identity.

$$\cos^{2}x=\frac{1+\cos2x}{2}.$$

Step 2: Substitute.

$$I=\int_{0}^{\pi/2}\frac{1+\cos2x}{2}\,dx = \frac12\int_{0}^{\pi/2}(1+\cos2x)\,dx.$$

Step 3: Integrate term by term. Using $$\int 1\,dx = x$$ and $$\int \cos2x\,dx = \dfrac{\sin2x}{2}$$:

$$I=\frac12\!\left[x+\frac{\sin2x}{2}\right]_{0}^{\pi/2}.$$

Step 4: Apply the limits. At $$x=\pi/2$$: $$\sin\pi = 0$$. At $$x=0$$: $$\sin0 = 0$$. So

$$I = \frac12\!\left[\!\left(\tfrac{\pi}{2}+0\right) - (0+0)\right] = \frac{\pi}{4}.$$

Step 1 : Identify a suitable substitution

The integrand contains the expression $$x^2+1$$ in the denominator and its derivative $$2x$$ (up to the constant 2) in the numerator. Thus, let us put

$$u = x^2 + 1$$

Then

$$\dfrac{du}{dx}=2x \;\;\Longrightarrow\;\; du = 2x\,dx \;\;\Longrightarrow\;\; x\,dx = \dfrac{1}{2}\,du$$

Step 2 : Change the limits

• When $$x = 2$$, $$u = 2^2 + 1 = 5$$
• When $$x = 3$$, $$u = 3^2 + 1 = 10$$

Step 3 : Rewrite and integrate

Using the substitution and the new limits:

$$\displaystyle\int_{2}^{3} \dfrac{x\,dx}{x^2+1} = \int_{u=5}^{u=10} \dfrac{\tfrac{1}{2}\,du}{u}$$

Extract the constant $$\tfrac12$$:

$$= \dfrac{1}{2}\,\int_{5}^{10} \dfrac{du}{u}$$

Step 4 : Evaluate the logarithmic integral

The antiderivative of $$\dfrac1u$$ is $$\ln|u|$$, so

$$\dfrac{1}{2}\Bigl[ \ln|u| \Bigr]_{5}^{10} = \dfrac{1}{2}\Bigl( \ln 10 - \ln 5 \Bigr)$$

Combine the logarithms:

$$\ln 10 - \ln 5 = \ln\!\left(\dfrac{10}{5}\right) = \ln 2$$

Step 5 : Final value

$$\displaystyle\int_{2}^{3} \dfrac{x\,dx}{x^2 + 1} = \dfrac{1}{2}\,\ln 2$$

We have to find

$$I = \int_{0}^{1}\dfrac{2x+3}{5x^{2}+1}\,dx$$

Split the integrand:

$$I = \int_{0}^{1}\dfrac{2x}{5x^{2}+1}\,dx + \int_{0}^{1}\dfrac{3}{5x^{2}+1}\,dx$$

Denote the two parts by $$I_{1}$$ and $$I_{2}$$, respectively.

1. Evaluation of $$I_{1}=\displaystyle\int_{0}^{1}\dfrac{2x}{5x^{2}+1}\,dx$$

Notice that the denominator’s derivative is proportional to the numerator:

$$\dfrac{d}{dx}(5x^{2}+1)=10x$$

Therefore, write

$$\dfrac{2x}{5x^{2}+1}=\dfrac{1}{5}\,\dfrac{10x}{5x^{2}+1}$$

Hence

$$I_{1}=\dfrac{1}{5}\int_{0}^{1}\dfrac{10x}{5x^{2}+1}\,dx=\dfrac{1}{5}\bigl[\ln(5x^{2}+1)\bigr]_{0}^{1}$$

Substituting the limits:

$$I_{1}=\dfrac{1}{5}\bigl(\ln 6-\ln 1\bigr)=\dfrac{1}{5}\ln 6$$

2. Evaluation of $$I_{2}=\displaystyle\int_{0}^{1}\dfrac{3}{5x^{2}+1}\,dx$$

Factor 3 out:

$$I_{2}=3\int_{0}^{1}\dfrac{dx}{5x^{2}+1}$$

Put $$x=\dfrac{t}{\sqrt5}\;\;\Longrightarrow\;\;dx=\dfrac{dt}{\sqrt5},\;5x^{2}+1=t^{2}+1$$

Thus

$$I_{2}=3\int_{0}^{\sqrt5}\dfrac{1}{\sqrt5}\,\dfrac{dt}{t^{2}+1}=\dfrac{3}{\sqrt5}\Bigl[\tan^{-1}t\Bigr]_{0}^{\sqrt5}$$

Evaluating:

$$I_{2}=\dfrac{3}{\sqrt5}\,\bigl(\tan^{-1}(\sqrt5)-\tan^{-1}0\bigr)=\dfrac{3}{\sqrt5}\tan^{-1}(\sqrt5)$$

3. Combine the two parts

$$I = I_{1}+I_{2}=\dfrac{1}{5}\ln 6+\dfrac{3}{\sqrt5}\tan^{-1}(\sqrt5)$$

Hence, the required definite integral equals

$$\boxed{\displaystyle\int_{0}^{1}\dfrac{2x+3}{5x^{2}+1}\,dx=\dfrac{1}{5}\ln 6+\dfrac{3}{\sqrt5}\tan^{-1}(\sqrt5)}$$

We must evaluate the definite integral $$\int_0^1 x e^{x^2}\,dx$$.

Step 1: Make a substitution.
Choose $$t = x^2$$ so that $$dt = 2x\,dx \Longrightarrow x\,dx = \dfrac{dt}{2}.$$

Step 2: Transform the limits.
When $$x = 0$$, $$t = 0$$; when $$x = 1$$, $$t = 1$$.

Step 3: Rewrite the integral.
$$\int_{0}^{1} x e^{x^2}\,dx = \int_{t=0}^{t=1} e^{t}\,\frac{dt}{2} = \frac12 \int_{0}^{1} e^{t}\,dt.$$

Step 4: Integrate and evaluate.
$$\frac12 \int_{0}^{1} e^{t}\,dt = \frac12 \bigl[e^{t}\bigr]_{0}^{1} = \frac12\bigl(e - 1\bigr).$$

Hence the required integral equals $$\dfrac{e-1}{2}$$.

Let
$$I=\int_1^2 \dfrac{5x^2}{x^2+4x+3}\,dx.$$

1. Divide the numerator by the denominator

Using polynomial (or long) division

We have to evaluate the definite integral

$$I = \displaystyle \int_{-1}^{1} 5x^4\sqrt{x^5+1}\,dx.$$

Step 1 : Choose a convenient substitution

The presence of both $$x^5$$ and its derivative $$5x^4$$ in the integrand suggests

$$u = x^5 + 1.$$

Differentiating,

$$du = 5x^4\,dx \quad\Longrightarrow\quad 5x^4\,dx = du.$$

Thus the product $$5x^4\,dx$$ in the integral can be replaced directly by $$du$$.

Step 2 : Change the limits of integration

• When $$x = -1$$: $$u = (-1)^5 + 1 = -1 + 1 = 0.$$
• When $$x = 1$$: $$u = 1^5 + 1 = 1 + 1 = 2.$$

So under the substitution, the integral becomes

$$I = \int_{0}^{2} \sqrt{u}\,du.$$

Step 3 : Integrate with respect to the new variable

Recall $$\sqrt{u} = u^{1/2}$$. Therefore,

$$I = \int_{0}^{2} u^{1/2}\,du = \Bigg[ \frac{u^{3/2}}{\tfrac{3}{2}} \Bigg]_{0}^{2} = \frac{2}{3}\,u^{3/2}\Big|_{0}^{2}.$$

Evaluating at the limits,

$$I = \frac{2}{3}\Bigl(2^{3/2} - 0^{3/2}\Bigr) = \frac{2}{3}\cdot 2^{3/2}.$$

Since $$2^{3/2} = 2^{1}\,2^{1/2} = 2\sqrt{2},$$ we get

$$I = \frac{2}{3}\times 2\sqrt{2} = \frac{4\sqrt{2}}{3}.$$

Therefore

$$\displaystyle \int_{-1}^{1} 5x^4\sqrt{x^5 + 1}\,dx = \frac{4\sqrt{2}}{3}. $$

Let
$$I = \int_{0}^{1} \dfrac{\tan^{-1}x}{1+x^{2}}\,dx$$

Step 1. Substitute $$x = \tan t$$.
Then $$t = \tan^{-1}x,$$ $$dx = \sec^{2}t\,dt,$$ and $$1+x^{2}=1+\tan^{2}t = \sec^{2}t.$$

Step 2. Transform the integrand:

• Numerator: $$\tan^{-1}x = t$$
• Denominator: $$1+x^{2} = \sec^{2}t$$
• Differential: $$dx = \sec^{2}t\,dt$$

Hence

$$\dfrac{\tan^{-1}x}{1+x^{2}}\,dx = \dfrac{t}{\sec^{2}t}\;\sec^{2}t\,dt = t\,dt.$$

Step 3. Change the limits:

• When $$x=0$$, $$t = \tan^{-1}0 = 0.$$
• When $$x=1$$, $$t = \tan^{-1}1 = \dfrac{\pi}{4}.$$

Step 4. Integrate:

$$I = \int_{0}^{\pi/4} t\,dt = \left[\dfrac{t^{2}}{2}\right]_{0}^{\pi/4} = \dfrac{1}{2}\left(\dfrac{\pi}{4}\right)^{2} = \dfrac{\pi^{2}}{32}.$$

Therefore,

$$\int_{0}^{1}\dfrac{\tan^{-1}x}{1+x^{2}}\,dx = \dfrac{\pi^{2}}{32}.$$

We must evaluate the definite integral

$$I = \displaystyle \int_0^1 \dfrac{x}{x^2 + 1}\,dx.$$

Step 1: Choose a suitable substitution.

Because the denominator is $$x^2 + 1$$ and its derivative is $$2x,$$ we let

$$t = x^2 + 1.$$

Then the differential is

$$dt = 2x\,dx \quad \Longrightarrow \quad x\,dx = \dfrac{dt}{2}.$$

Step 2: Transform the limits of integration.

• When $$x = 0$$, $$t = 0^2 + 1 = 1.$$
• When $$x = 1$$, $$t = 1^2 + 1 = 2.$$

Step 3: Rewrite the integral in terms of $$t$$.

Substituting $$t$$ and $$x\,dx$$, we get

$$I = \int_{t=1}^{t=2} \dfrac{1}{t}\,\frac{dt}{2} = \dfrac{1}{2}\int_{1}^{2} \dfrac{dt}{t}.$$

Step 4: Integrate.

The antiderivative of $$1/t$$ is $$\ln t$$, so

$$I = \dfrac{1}{2}\Bigl[\ln t\Bigr]_{1}^{2} = \dfrac{1}{2}\bigl(\ln 2 - \ln 1\bigr).$$

Step 5: Simplify.

Since $$\ln 1 = 0$$, we obtain

$$I = \dfrac{1}{2}\ln 2.$$

Hence, the value of the integral is $$\dfrac{1}{2}\ln 2.$$

We have to evaluate the definite integral

$$I = \displaystyle\int_{0}^{\pi/2} \sqrt{\sin \phi}\,\cos^{5}\phi\,d\phi.$$

Step 1: Substitution. Put $$t = \sin \phi$$, so $$dt = \cos\phi\,d\phi.$$ Limits: $$\phi=0 \Rightarrow t=0$$, $$\phi=\tfrac\pi2 \Rightarrow t=1.$$

Step 2: Transform the integrand. Write $$\cos^{5}\phi\,d\phi = \cos^{4}\phi \cdot \cos\phi\,d\phi = \cos^{4}\phi\,dt.$$ Since $$\cos^{2}\phi = 1-\sin^{2}\phi = 1-t^{2}$$, we have $$\cos^{4}\phi = (1-t^{2})^{2}.$$ Therefore

$$I = \int_{0}^{1} t^{1/2}\,(1-t^{2})^{2}\,dt.$$

Step 3: Expand and integrate.

$$(1-t^{2})^{2} = 1 - 2t^{2} + t^{4},$$

so

$$I = \int_{0}^{1}\!\left(t^{1/2} - 2t^{5/2} + t^{9/2}\right)dt = \left[\frac{2}{3}t^{3/2} - \frac{4}{7}t^{7/2} + \frac{2}{11}t^{11/2}\right]_{0}^{1} = \frac{2}{3} - \frac{4}{7} + \frac{2}{11}.$$

Step 4: Simplify. Common denominator $$231$$:

$$I = \frac{154 - 132 + 42}{231} = \frac{64}{231}.$$

Let

$$I = \int_{0}^{1} \sin^{-1}\!\left(\dfrac{2x}{1 + x^{2}}\right)\,dx.$$

Step 1: Trigonometric substitution

Put $$x = \tan\theta \;(0\le x\le 1)\;\Rightarrow\;\theta = \tan^{-1}x,$$ so that $$dx = \sec^{2}\theta\,d\theta.$$

Then

$$\dfrac{2x}{1+x^{2}} = \dfrac{2\tan\theta}{1+\tan^{2}\theta}=\dfrac{2\tan\theta}{\sec^{2}\theta}=2\tan\theta\cos^{2}\theta=2\sin\theta\cos\theta = \sin 2\theta.$$

When $$x$$ increases from $$0$$ to $$1$$, $$\theta$$ increases from $$0$$ to $$\dfrac{\pi}{4}$$, so $$0\le 2\theta\le\dfrac{\pi}{2}$$. In this interval the principal value satisfies $$\sin^{-1}(\sin 2\theta)=2\theta.$$

Step 2: Change of variable

Hence

$$I = \int_{0}^{\pi/4} 2\theta\,\sec^{2}\theta\,d\theta = 2\int_{0}^{\pi/4} \theta\sec^{2}\theta\,d\theta.$$

Step 3: Integration by parts

Take $$u = \theta \;\Rightarrow\;du = d\theta,$$ $$dv = \sec^{2}\theta\,d\theta \;\Rightarrow\;v = \tan\theta.$$

Then

$$I = 2\Big[\theta\tan\theta\Big]_{0}^{\pi/4} - 2\int_{0}^{\pi/4}\tan\theta\,d\theta.$$

Evaluate the boundary term:

$$\Big[\theta\tan\theta\Big]_{0}^{\pi/4}=\left(\dfrac{\pi}{4}\cdot 1\right)-0=\dfrac{\pi}{4}.$$

The remaining integral:

$$\int\tan\theta\,d\theta = -\ln|\cos\theta|.$$

Hence

$$I = 2\cdot\dfrac{\pi}{4}-2\Big[-\ln|\cos\theta|\Big]_{0}^{\pi/4}=\dfrac{\pi}{2}+2\Big[\ln|\cos\theta|\Big]_{0}^{\pi/4}.$$

Compute the logarithmic term:

$$\cos\dfrac{\pi}{4}=\dfrac{\sqrt{2}}{2}=\dfrac{1}{\sqrt{2}}, \qquad \cos 0 = 1,$$

so

$$\ln|\cos\dfrac{\pi}{4}|-\ln|\cos 0| = \ln\left(\dfrac{1}{\sqrt{2}}\right)-\ln 1 = -\dfrac{1}{2}\ln 2.$$

Therefore

$$I = \dfrac{\pi}{2} + 2\left(-\dfrac{1}{2}\ln 2\right)=\dfrac{\pi}{2}-\ln 2.$$

Hence

$$\displaystyle\int_{0}^{1} \sin^{-1}\!\left(\dfrac{2x}{1 + x^{2}}\right) dx = \dfrac{\pi}{2}-\ln 2.$$

We have to evaluate the definite integral

$$I = \displaystyle\int_{0}^{2} x\sqrt{x+2}\,dx$$

Put the suggested substitution

$$x+2 = t^{2}\;\;(t>0) \;\;\Longrightarrow\;\; x = t^{2}-2, \; dx = 2t\,dt$$

New limits

• When $$x = 0$$, $$t^{2} = 0 + 2 \Rightarrow t = \sqrt{2}$$
• When $$x = 2$$, $$t^{2} = 2 + 2 \Rightarrow t = 2$$

Hence

$$I = \int_{t = \sqrt{2}}^{2} \bigl(t^{2}-2\bigr)\,\sqrt{t^{2}}\,(2t)\,dt$$

Since $$t>0$$ in the interval $$[\sqrt{2},2]$$, $$\sqrt{t^{2}} = t$$. Therefore

$$I = \int_{\sqrt{2}}^{2} \bigl(t^{2}-2\bigr)\,t\,(2t)\,dt = \int_{\sqrt{2}}^{2} 2\bigl(t^{2}-2\bigr)t^{2}\,dt$$

Expand the bracket:

$$I = 2\int_{\sqrt{2}}^{2} \bigl(t^{4}-2t^{2}\bigr)\,dt$$

Integrate term-wise:

$$I = 2\left[\frac{t^{5}}{5}-2\cdot\frac{t^{3}}{3}\right]_{\sqrt{2}}^{2} = 2\left[\frac{t^{5}}{5}-\frac{2t^{3}}{3}\right]_{\sqrt{2}}^{2}$$

Upper limit $$t=2$$:

$$\frac{2^{5}}{5}-\frac{2\,2^{3}}{3}=\frac{32}{5}-\frac{16}{3}$$

Lower limit $$t=\sqrt{2}$$:

$$\frac{(\sqrt{2})^{5}}{5}-\frac{2\,(\sqrt{2})^{3}}{3} = \frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}$$

Subtract and multiply by the outer 2:

$$I = 2\Bigl[\left(\frac{32}{5}-\frac{16}{3}\right)-\left(\frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}\right)\Bigr]$$

Simplify the rational part:

$$\frac{32}{5}-\frac{16}{3}=\frac{96-80}{15}=\frac{16}{15}$$

Simplify the surd part:

$$\frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}=4\sqrt{2}\left(\frac{1}{5}-\frac{1}{3}\right) =4\sqrt{2}\left(-\frac{2}{15}\right)=-\frac{8\sqrt{2}}{15}$$

Hence the bracket equals

$$\frac{16}{15}-\left(-\frac{8\sqrt{2}}{15}\right)=\frac{16+8\sqrt{2}}{15}$$

Finally

$$I = 2\cdot\frac{16+8\sqrt{2}}{15}=\frac{32+16\sqrt{2}}{15}=\frac{16\,(2+\sqrt{2})}{15}$$

Therefore, the value of the integral is

$$\boxed{\dfrac{16\,(2+\sqrt{2})}{15}}$$

We have to evaluate the definite integral

$$I = \displaystyle \int_{0}^{\pi/2} \dfrac{\sin x}{1 + \cos^{2}x}\,dx$$

Step 1 – Choose a suitable substitution

The integrand contains both $$\sin x$$ and $$\cos x$$, with $$\cos x$$ appearing inside a simple algebraic expression. A natural choice is

$$u = \cos x$$

Then

$$\dfrac{du}{dx} = -\sin x \;\;\;\Longrightarrow\;\;\; du = -\sin x\,dx \;\;\;\Longrightarrow\;\;\; \sin x\,dx = -du$$

Step 2 – Change the limits of integration

The original limits are values of $$x$$; we convert them to limits for $$u$$:

• When $$x = 0$$, $$u = \cos 0 = 1$$.
• When $$x = \dfrac{\pi}{2}$$, $$u = \cos \dfrac{\pi}{2} = 0$$.

Step 3 – Rewrite the integral in terms of $$u$$

Substituting $$u$$ and $$du$$, we get

$$I = \int_{x = 0}^{x = \pi/2} \dfrac{\sin x}{1 + \cos^{2}x}\,dx = \int_{u = 1}^{u = 0} \dfrac{-du}{1 + u^{2}}$$

Factor the minus sign and switch the limits (which removes the minus sign):

$$I = - \int_{1}^{0} \dfrac{du}{1 + u^{2}} = \int_{0}^{1} \dfrac{du}{1 + u^{2}}$$

Step 4 – Integrate

For any real number $$u$$,

$$\int \dfrac{du}{1 + u^{2}} = \tan^{-1} u + C$$

Therefore,

$$I = \left[\tan^{-1} u\right]_{0}^{1} = \tan^{-1} 1 \; - \; \tan^{-1} 0$$

Since $$\tan^{-1} 1 = \dfrac{\pi}{4}$$ and $$\tan^{-1} 0 = 0$$,

$$I = \dfrac{\pi}{4} - 0 = \dfrac{\pi}{4}$$

Conclusion

$$\displaystyle \int_{0}^{\pi/2} \dfrac{\sin x}{1 + \cos^{2}x}\,dx = \dfrac{\pi}{4}$$

We have to evaluate

$$I=\int_{0}^{2}\dfrac{dx}{x+4-x^{2}}.$$

1. Rewrite the denominator. Factor out the minus sign and complete the square:

$$x+4-x^{2}=-(x^{2}-x-4) = -\!\left[\left(x-\tfrac12\right)^{2}-\tfrac{17}{4}\right] = \tfrac{17}{4}-\left(x-\tfrac12\right)^{2}.$$

2. Substitution. Put $$u=x-\tfrac12$$, so $$du=dx$$ and the limits change to $$u=-\tfrac12$$ (when $$x=0$$) and $$u=\tfrac32$$ (when $$x=2$$). Then

$$I=\int_{-1/2}^{3/2}\dfrac{du}{\tfrac{17}{4}-u^{2}}.$$

3. Apply the standard form $$\displaystyle\int\dfrac{du}{a^{2}-u^{2}}=\dfrac{1}{2a}\ln\!\left|\dfrac{a+u}{a-u}\right|+C$$ (valid for $$a>0$$), with $$a=\tfrac{\sqrt{17}}{2}$$, $$\tfrac{1}{2a}=\tfrac{1}{\sqrt{17}}$$:

$$I=\tfrac{1}{\sqrt{17}}\!\left[\ln\!\left|\dfrac{a+u}{a-u}\right|\right]_{-1/2}^{3/2}.$$

4. Substitute the limits.

• At $$u=\tfrac32$$: $$\dfrac{a+u}{a-u}=\dfrac{\sqrt{17}+3}{\sqrt{17}-3}.$$
• At $$u=-\tfrac12$$: $$\dfrac{a+u}{a-u}=\dfrac{\sqrt{17}-1}{\sqrt{17}+1}.$$

Hence

$$I=\tfrac{1}{\sqrt{17}}\!\left[\ln\!\left(\dfrac{\sqrt{17}+3}{\sqrt{17}-3}\right) - \ln\!\left(\dfrac{\sqrt{17}-1}{\sqrt{17}+1}\right)\right] = \tfrac{1}{\sqrt{17}}\ln\!\left(\dfrac{(\sqrt{17}+3)(\sqrt{17}+1)}{(\sqrt{17}-3)(\sqrt{17}-1)}\right).$$

5. Simplify.

$$(\sqrt{17}+3)(\sqrt{17}+1) = 17 + 4\sqrt{17} + 3 = 20 + 4\sqrt{17} = 4(5+\sqrt{17}),$$ $$(\sqrt{17}-3)(\sqrt{17}-1) = 17 - 4\sqrt{17} + 3 = 20 - 4\sqrt{17} = 4(5-\sqrt{17}).$$

The factor $$4$$ cancels, giving

$$I=\tfrac{1}{\sqrt{17}}\ln\!\left(\dfrac{5+\sqrt{17}}{5-\sqrt{17}}\right).$$

We have to evaluate

$$I = \displaystyle \int_{-1}^{1} \frac{dx}{x^{2}+2x+5}.$$

Step 1: Complete the square. $$x^{2}+2x+5 = (x+1)^{2}+4.$$ Thus

$$I = \int_{-1}^{1} \frac{dx}{(x+1)^{2}+4}.$$

Step 2: Substitute $$t = x+1$$, so $$dt=dx$$ and the limits become $$t=0$$ (when $$x=-1$$) and $$t=2$$ (when $$x=1$$):

$$I = \int_{0}^{2} \frac{dt}{t^{2}+4}.$$

Step 3: Apply the standard formula $$\int \dfrac{du}{u^{2}+a^{2}} = \dfrac{1}{a}\tan^{-1}\!\left(\dfrac{u}{a}\right) + C$$ with $$a=2$$:

$$I = \left[\tfrac12\tan^{-1}\!\left(\tfrac{t}{2}\right)\right]_{0}^{2}.$$

Step 4: Evaluate.

$$I = \tfrac12\!\left[\tan^{-1}(1) - \tan^{-1}(0)\right] = \tfrac12 \cdot \tfrac{\pi}{4} = \tfrac{\pi}{8}.$$

Let

$$I = \int_{1}^{2} \left(\frac{1}{x}-\frac{1}{2x^{2}}\right)e^{2x}\,dx.$$

Rewrite the bracket so that a familiar derivative appears:

$$\frac{1}{x}-\frac{1}{2x^{2}} = \tfrac12\!\left(\frac{2}{x}-\frac{1}{x^{2}}\right).$$

Hence

$$I = \tfrac12 \int_{1}^{2}\!\left(\frac{2}{x}-\frac{1}{x^{2}}\right)e^{2x}\,dx.$$

Observe that the derivative of $$\dfrac{e^{2x}}{x}$$ is exactly the integrand inside this integral:

$$\frac{d}{dx}\!\left(\frac{e^{2x}}{x}\right)=\frac{x\cdot 2e^{2x}-e^{2x}}{x^{2}}=e^{2x}\!\left(\frac{2}{x}-\frac{1}{x^{2}}\right).$$

Therefore

$$\left(\frac{2}{x}-\frac{1}{x^{2}}\right)e^{2x}\,dx = d\!\left(\frac{e^{2x}}{x}\right),$$

and

$$I = \tfrac12 \left[\frac{e^{2x}}{x}\right]_{1}^{2} = \tfrac12\!\left(\frac{e^{4}}{2}-\frac{e^{2}}{1}\right) = \frac{e^{4}}{4}-\frac{e^{2}}{2} = \frac{e^{2}(e^{2}-2)}{4}.$$

The integral to be evaluated is

$$I = \displaystyle\int_{1/3}^{1} \frac{(x-x^{3})^{1/3}}{x^{4}}\,dx.$$

Factor under the cube root: $$x-x^{3}=x(1-x^{2})$$, so

$$I = \int_{1/3}^{1}\frac{\bigl[x(1-x^{2})\bigr]^{1/3}}{x^{4}}\,dx = \int_{1/3}^{1}x^{-11/3}(1-x^{2})^{1/3}\,dx.$$

Rewrite the integrand by pulling $$x^{3}$$ out of the cube root: $$\dfrac{(x-x^{3})^{1/3}}{x^{4}} = \dfrac{x^{3\cdot 1/3}\!\left(\dfrac{1}{x^{2}}-1\right)^{1/3}}{x^{4}} = \dfrac{1}{x^{3}}\!\left(\dfrac{1}{x^{2}}-1\right)^{1/3}.$$ Put $$u = \dfrac{1}{x^{2}} - 1$$, so $$du = -\dfrac{2}{x^{3}}\,dx \Rightarrow \dfrac{dx}{x^{3}} = -\dfrac{du}{2}.$$

Limits: when $$x = 1$$, $$u = 0$$; when $$x = \tfrac13$$, $$u = 9-1 = 8.$$

Hence

$$I = \int_{u=8}^{u=0} u^{1/3}\!\left(-\tfrac12\,du\right) = \tfrac12 \int_{0}^{8} u^{1/3}\,du = \tfrac12 \cdot \tfrac34 \bigl[u^{4/3}\bigr]_{0}^{8}.$$

Since $$8^{4/3} = (8^{1/3})^{4} = 2^{4} = 16$$,

$$I = \tfrac12 \cdot \tfrac34 \cdot 16 = 6.$$

Hence the correct option is (A) 6.

Given the function

$$f(x)=\int_0^x t\sin t\,dt,$$

observe that the integrand $$g(t)=t\sin t$$ is a product of two continuous functions on \([0, x]\). Hence $$g(t)$$ is continuous on any closed interval contained in \([0,\infty)\).

The Fundamental Theorem of Calculus – Part I states:

If $$G(x)=\int_a^x g(t)\,dt$$ with $$g(t)$$ continuous on an interval containing $$a$$ and $$x$$, then

$$G'(x)=g(x).$$

Here,

• $$g(t)=t\sin t,$$
• lower limit $$a=0,$$
• upper limit $$x$$ is the variable.

Applying the theorem directly, we obtain

$$f'(x)=g(x)=x\sin x.$$

Among the given alternatives, this matches option (B).

Let $$f(x)=x^{3}-x=x(x-1)(x+1).$$ The zeros of $$f$$ are $$x=-1,\;0,\;1$$, which split $$[-1,2]$$ into three sub-intervals. Determine the sign of $$f$$ on each sub-interval using a test point:

• On $$[-1,0]$$: at $$x=-\tfrac12$$, $$f\!\left(-\tfrac12\right)=-\tfrac18+\tfrac12=\tfrac{3}{8}.$$ Since $$\tfrac{3}{8}\gt 0$$, we have $$f(x)\ge 0$$ on $$[-1,0]$$.
• On $$[0,1]$$: at $$x=\tfrac12$$, $$f\!\left(\tfrac12\right)=\tfrac18-\tfrac12=-\tfrac{3}{8}.$$ Since $$-\tfrac{3}{8}\lt 0$$, we have $$f(x)\le 0$$ on $$[0,1]$$.
• On $$[1,2]$$: at $$x=\tfrac32$$, $$f\!\left(\tfrac32\right)=\tfrac{27}{8}-\tfrac{12}{8}=\tfrac{15}{8}.$$ Since $$\tfrac{15}{8}\gt 0$$, we have $$f(x)\ge 0$$ on $$[1,2]$$.

Hence

$$|x^{3}-x|=\begin{cases}x^{3}-x,&-1\le x\le 0,\\ -(x^{3}-x),&0\le x\le 1,\\ x^{3}-x,&1\le x\le 2.\end{cases}$$

Therefore

$$\int_{-1}^{2}|x^{3}-x|\,dx=\int_{-1}^{0}(x^{3}-x)\,dx-\int_{0}^{1}(x^{3}-x)\,dx+\int_{1}^{2}(x^{3}-x)\,dx.$$

An antiderivative of $$x^{3}-x$$ is $$F(x)=\dfrac{x^{4}}{4}-\dfrac{x^{2}}{2}.$$ Evaluating each piece:

• $$\displaystyle\int_{-1}^{0}(x^{3}-x)\,dx=F(0)-F(-1)=0-\left(\tfrac14-\tfrac12\right)=\tfrac14.$$
• $$\displaystyle\int_{0}^{1}(x^{3}-x)\,dx=F(1)-F(0)=\left(\tfrac14-\tfrac12\right)-0=-\tfrac14,$$ so $$\displaystyle-\int_{0}^{1}(x^{3}-x)\,dx=\tfrac14.$$
• $$\displaystyle\int_{1}^{2}(x^{3}-x)\,dx=F(2)-F(1)=(4-2)-\left(\tfrac14-\tfrac12\right)=2+\tfrac14=\tfrac94.$$

Adding the three contributions,

$$\int_{-1}^{2}|x^{3}-x|\,dx=\tfrac14+\tfrac14+\tfrac94=\dfrac{11}{4}.$$

We have to evaluate the definite integral

$$I = \displaystyle\int_{-\pi/4}^{\pi/4} \sin^2 x\,dx.$$

1. Use the symmetry of the integrand. Since $$\sin^2(-x) = \sin^2 x$$, $$\sin^2 x$$ is even, so

$$I = 2 \int_{0}^{\pi/4} \sin^2 x\,dx.$$

2. Convert $$\sin^2 x$$ via the power-reducing identity $$\sin^2 x = \dfrac{1-\cos 2x}{2}$$:

$$I = 2 \int_{0}^{\pi/4} \frac{1-\cos 2x}{2}\,dx = \int_{0}^{\pi/4} (1-\cos 2x)\,dx.$$

3. Integrate term by term.

• $$\int_{0}^{\pi/4} 1\,dx = \tfrac{\pi}{4}.$$
• $$\int_{0}^{\pi/4} \cos 2x\,dx = \left[\tfrac{\sin 2x}{2}\right]_{0}^{\pi/4} = \tfrac{\sin(\pi/2)}{2} = \tfrac{1}{2}.$$

Therefore

$$I = \frac{\pi}{4} - \frac{1}{2}.$$

Let $$I = \displaystyle\int_0^{\pi} \dfrac{x\,\sin x}{1+\cos^2 x}\,dx$$.

Step 1 – Use the property $$\displaystyle\int_0^{a}f(x)\,dx=\int_0^{a}f(a-x)\,dx.$$

Put $$x=\pi-t \;(t=\pi-x)\Rightarrow dx=-dt.$$ Then

$$I = \int_{x=0}^{x=\pi}\dfrac{x\sin x}{1+\cos^2 x}\,dx = \int_{t=\pi}^{t=0}\dfrac{(\pi-t)\sin(\pi-t)}{1+\cos^2(\pi-t)}(-dt).$$

Because $$\sin(\pi-t)=\sin t$$ and $$\cos(\pi-t)= -\cos t,$$ we get

$$I = \int_{0}^{\pi}\dfrac{(\pi-t)\sin t}{1+\cos^2 t}\,dt.$$

Step 2 – Add the two expressions for $$I.$$

Original integral:
$$I = \int_0^{\pi}\dfrac{x\sin x}{1+\cos^2 x}\,dx.$$

Transformed integral:
$$I = \int_0^{\pi}\dfrac{(\pi-x)\sin x}{1+\cos^2 x}\,dx.$$

Adding,

$$2I = \int_0^{\pi}\dfrac{\bigl[x+(\pi-x)\bigr]\sin x}{1+\cos^2 x}\,dx = \pi\int_0^{\pi}\dfrac{\sin x}{1+\cos^2 x}\,dx.$$

Hence

$$I = \dfrac{\pi}{2}\,K \quad \text{where} \quad K = \int_0^{\pi}\dfrac{\sin x}{1+\cos^2 x}\,dx.$$

Step 3 – Evaluate $$K.$$

Put $$t=\cos x \;\;(dt=-\sin x\,dx).$$ Then $$\sin x\,dx = -dt.$$

Limits: when $$x=0,\;t=\cos0=1;\qquad x=\pi,\;t=\cos\pi=-1.$$

Therefore

$$K = \int_{1}^{-1}\dfrac{-dt}{1+t^{2}} = \int_{-1}^{1}\dfrac{dt}{1+t^{2}}.$$

The antiderivative of $$\dfrac{1}{1+t^{2}}$$ is $$\tan^{-1}t$$, so

$$K = \bigl[\tan^{-1}t\bigr]_{-1}^{1} = \tan^{-1}(1)-\tan^{-1}(-1)=\dfrac{\pi}{4}-\left(-\dfrac{\pi}{4}\right)=\dfrac{\pi}{2}.$$

Step 4 – Substitute back into $$I.$$

$$I = \dfrac{\pi}{2}\times\dfrac{\pi}{2}=\dfrac{\pi^{2}}{4}.$$

Therefore

$$\displaystyle\int_0^{\pi}\dfrac{x\sin x}{1+\cos^{2}x}\,dx = \dfrac{\pi^{2}}{4}.$$

Let

$$I = \int_{-1}^{1} \sin^{5}x\,\cos^{4}x\,dx.$$

Define $$f(x)=\sin^{5}x\,\cos^{4}x.$$

Step 1: Check parity. For any real $$x$$,

$$f(-x)=\sin^{5}(-x)\,\cos^{4}(-x)=(-\sin x)^{5}(\cos x)^{4}=-\sin^{5}x\,\cos^{4}x=-f(x).$$

So $$f$$ is an odd function.

Step 2: Use the property of odd integrals. Since $$f$$ is continuous and odd on $$[-1,1]$$,

$$I = \int_{-1}^{1} \sin^{5}x\,\cos^{4}x\,dx = 0.$$

Let

$$I = \displaystyle \int_0^{\pi/2} \frac{\sin^4 x}{\sin^4 x + \cos^4 x}\,dx$$

Step 1 : Apply the property $$\displaystyle\int_0^{a} f(x)\,dx = \int_0^{a} f(a-x)\,dx$$

Put $$x = \frac{\pi}{2} - t \;\; (\Rightarrow\; dx = -dt)$$ in the definition of $$I$$:

$$I = \int_{x=0}^{x=\pi/2} \frac{\sin^4 x}{\sin^4 x + \cos^4 x}\,dx = \int_{t=\pi/2}^{t=0} \frac{\sin^4\bigl(\tfrac{\pi}{2}-t\bigr)}{\sin^4\bigl(\tfrac{\pi}{2}-t\bigr)+\cos^4\bigl(\tfrac{\pi}{2}-t\bigr)}(-dt)$$

Because $$\sin\bigl(\tfrac{\pi}{2}-t\bigr)=\cos t$$ and $$\cos\bigl(\tfrac{\pi}{2}-t\bigr)=\sin t$$, this becomes

$$I = \int_{0}^{\pi/2} \frac{\cos^4 t}{\cos^4 t + \sin^4 t}\,dt$$

Rename the dummy variable $$t$$ back to $$x$$:

$$I = \int_{0}^{\pi/2} \frac{\cos^4 x}{\sin^4 x + \cos^4 x}\,dx$$

Step 2 : Add the two forms of $$I$$

Write the original and transformed versions together:

$$I = \int_{0}^{\pi/2} \frac{\sin^4 x}{\sin^4 x + \cos^4 x}\,dx$$

$$I = \int_{0}^{\pi/2} \frac{\cos^4 x}{\sin^4 x + \cos^4 x}\,dx$$

Add them:

$$2I = \int_{0}^{\pi/2} \Bigl( \frac{\sin^4 x}{\sin^4 x + \cos^4 x} + \frac{\cos^4 x}{\sin^4 x + \cos^4 x} \Bigr) dx$$

Simplify the integrand:

$$\frac{\sin^4 x + \cos^4 x}{\sin^4 x + \cos^4 x} = 1$$

Hence

$$2I = \int_{0}^{\pi/2} 1\,dx = \Bigl[x\Bigr]_0^{\pi/2} = \frac{\pi}{2}$$

Step 3 : Solve for $$I$$

$$2I = \frac{\pi}{2} \;\;\Longrightarrow\;\; I = \frac{\pi}{4}$$

Therefore,

$$\displaystyle \int_0^{\pi/2} \frac{\sin^4 x}{\sin^4 x + \cos^4 x}\,dx = \frac{\pi}{4}$$

Let
$$I = \int_{\pi/6}^{\pi/3}\dfrac{dx}{1+\sqrt{\tan x}}.$$

Notice that the end–points satisfy $$\dfrac{\pi}{6}+\dfrac{\pi}{3}=\dfrac{\pi}{2}.$$ Hence put $$x=\dfrac{\pi}{2}-t\;\;(\Rightarrow\;dx=-dt).$$ When $$x=\dfrac{\pi}{6},\;t=\dfrac{\pi}{3};$$ and when $$x=\dfrac{\pi}{3},\;t=\dfrac{\pi}{6}.$$ Re-writing the same integral with this change of variable we get

$$I = \int_{\pi/3}^{\pi/6}\dfrac{-dt}{1+\sqrt{\tan\bigl(\frac{\pi}{2}-t\bigr)}} = \int_{\pi/6}^{\pi/3}\dfrac{dt}{1+\sqrt{\cot t}}.$$ (Renaming the dummy variable $$t$$ back to $$x$$ gives the same limits.) Thus

$$I = \int_{\pi/6}^{\pi/3}\dfrac{dx}{1+\sqrt{\cot x}}.$$

Add the two equal expressions for $$I$$:

$$2I = \int_{\pi/6}^{\pi/3}\left[\dfrac{1}{1+\sqrt{\tan x}} + \dfrac{1}{1+\sqrt{\cot x}}\right]dx.$$

Write $$u=\sqrt{\tan x}\;\;(u>0 \text{ on the interval}).$$ Then $$\sqrt{\cot x}=\dfrac{1}{u}.$$ Inside the bracket we have

$$\dfrac{1}{1+u}+\dfrac{1}{1+\dfrac{1}{u}} \;=\; \dfrac{1}{1+u}+\dfrac{u}{u+1} = \dfrac{1+u}{1+u}=1.$$

Therefore the integrand in the sum is identically $$1$$, so

$$2I = \int_{\pi/6}^{\pi/3} 1\,dx = \left[x\right]_{\pi/6}^{\pi/3}=\dfrac{\pi}{3}-\dfrac{\pi}{6}=\dfrac{\pi}{6}.$$

Finally,

$$I = \dfrac{1}{2}\cdot\dfrac{\pi}{6}=\dfrac{\pi}{12}.$$

Hence the required value is $$\boxed{\dfrac{\pi}{12}}$$.

Let $$I = \displaystyle \int_{0}^{\pi/2} \log \sin x\,dx$$.

Step 1 : Show that the same integral occurs with cos x.

Put $$x = \dfrac{\pi}{2}-t$$. Then $$dx = -dt$$, and when $$x = 0 \;(\text{lower limit})$$ we have $$t = \dfrac{\pi}{2}$$, while when $$x = \dfrac{\pi}{2} \;(\text{upper limit})$$ we have $$t = 0$$. Therefore

$$I = \int_{\pi/2}^{0} \log \sin\left(\dfrac{\pi}{2}-t\right)(-dt)=\int_{0}^{\pi/2} \log \cos t\,dt.$$

Hence

$$I = \int_{0}^{\pi/2} \log \cos x\,dx.$$ Because these two integrals are equal, we may add them.

Step 2 : Add the two equal expressions.

$$2I = \int_{0}^{\pi/2}\bigl[\log \sin x + \log \cos x\bigr]dx = \int_{0}^{\pi/2}\log(\sin x\,\cos x)\,dx.$$

The product inside the logarithm can be rewritten using the double–angle identity $$\sin x\,\cos x = \tfrac12\sin 2x$$, so

$$2I = \int_{0}^{\pi/2} \log\Bigl(\tfrac12\sin 2x\Bigr)\,dx = \int_{0}^{\pi/2} \log \sin 2x\,dx + \int_{0}^{\pi/2} \log \tfrac12\,dx.$$ The second integral is a constant:

$$\int_{0}^{\pi/2}\log \tfrac12\,dx = \Bigl(\frac{\pi}{2}\Bigr)\log \tfrac12 = -\frac{\pi}{2}\log 2.$$ So

$$2I = \int_{0}^{\pi/2} \log \sin 2x\,dx - \frac{\pi}{2}\log 2.$$

Step 3 : Reduce the remaining integral to $I$ again.

Put $$t = 2x\;(\Rightarrow\;dt = 2dx).$$ When $$x$$ runs from 0 to $$\pi/2$$, $$t$$ runs from 0 to $$\pi$$. Hence

$$\int_{0}^{\pi/2} \log \sin 2x\,dx = \tfrac12\int_{0}^{\pi} \log \sin t\,dt.$$

Break the range $$(0,\pi)$$ into two halves:

$$\int_{0}^{\pi} \log \sin t\,dt = \int_{0}^{\pi/2} \log \sin t\,dt + \int_{\pi/2}^{\pi} \log \sin t\,dt.$$ In the second integral substitute $$t = \pi-u$$; then $$\sin t=\sin u$$ and the limits change from $$t: \pi/2 \to \pi$$ to $$u: \pi/2 \to 0$$, giving

$$\int_{\pi/2}^{\pi} \log \sin t\,dt = \int_{\pi/2}^{0} \log \sin u\,(-du)=\int_{0}^{\pi/2} \log \sin u\,du.$$

Thus the two halves are equal, so

$$\int_{0}^{\pi} \log \sin t\,dt = 2\int_{0}^{\pi/2} \log \sin t\,dt = 2I.$$

Consequently

$$\int_{0}^{\pi/2} \log \sin 2x\,dx = \tfrac12(2I)=I.$$

Step 4 : Collect the results and solve for $I$.

Substituting this back into the equation from Step 2:

$$2I = I - \frac{\pi}{2}\log 2.$$

Move $I$ to the left:

$$I = -\frac{\pi}{2}\log 2.$$

Therefore,

$$\boxed{\displaystyle \int_{0}^{\pi/2} \log \sin x\,dx = -\frac{\pi}{2}\log 2}.$$

Let $$I = \displaystyle\int_{0}^{\pi/2}\cos^{2}x\,dx.$$

1. Apply the property $$\displaystyle\int_{0}^{a}f(x)\,dx = \int_{0}^{a}f(a-x)\,dx$$ with $$f(x)=\cos^{2}x$$ and $$a=\tfrac{\pi}{2}$$:

$$I = \int_{0}^{\pi/2}\cos^{2}\!\Bigl(\tfrac{\pi}{2}-x\Bigr)\,dx = \int_{0}^{\pi/2}\sin^{2}x\,dx.$$

2. Add the two equal integrals:

$$2I = \int_{0}^{\pi/2}(\cos^{2}x + \sin^{2}x)\,dx = \int_{0}^{\pi/2} 1\,dx = \tfrac{\pi}{2}.$$

3. Solve for $$I$$:

$$I = \tfrac{\pi}{4}.$$

Let

$$I = \int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx$$

We employ the definite–integral property

$$\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx.$$

Take $$a = \frac{\pi}{2}$$ and define

$$f(x)=\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}.$$

Using the property,

$$I = \int_{0}^{\pi/2} f\!\left(\frac{\pi}{2}-x\right)dx.$$

Now compute $$f\!\left(\frac{\pi}{2}-x\right):$$

• $$\sin\!\left(\frac{\pi}{2}-x\right)=\cos x,$$
• $$\cos\!\left(\frac{\pi}{2}-x\right)=\sin x.$$

Thus

$$f\!\left(\frac{\pi}{2}-x\right)=\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}.$$

Hence we have two equal representations of $$I$$:

$$I = \int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx$$

and

$$I = \int_{0}^{\pi/2}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx.$$

Add the two expressions term-by-term:

$$2I = \int_{0}^{\pi/2}\left[\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\right]dx.$$

Both fractions have the same denominator, so add the numerators:

$$\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} = 1.$$

Therefore, the integrand simplifies to $$1$$ over the entire interval:

$$2I = \int_{0}^{\pi/2} 1\,dx = \left[x\right]_{0}^{\pi/2} = \frac{\pi}{2}.$$

Divide by $$2$$ to obtain $$I$$:

$$I = \frac{1}{2}\cdot\frac{\pi}{2} = \frac{\pi}{4}.$$

Hence

$$\int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx = \frac{\pi}{4}. $$

Let

$$I = \int_0^{\pi/2}\dfrac{\sin^{3/2}x\,dx}{\sin^{3/2}x+\cos^{3/2}x}$$

Step 1 : Apply the property

The definite-integral property $$\displaystyle\int_0^{a}f(x)\,dx = \int_0^{a}f(a-x)\,dx$$ is valid whenever the integral exists. Here $$a=\tfrac{\pi}{2}$$, so substitute $$x\;\to\;\tfrac{\pi}{2}-x$$ in the integrand:

$$I = \int_0^{\pi/2} \dfrac{\sin^{3/2}\bigl(\tfrac{\pi}{2}-x\bigr)\,dx}{\sin^{3/2}\bigl(\tfrac{\pi}{2}-x\bigr)+\cos^{3/2}\bigl(\tfrac{\pi}{2}-x\bigr)}$$

Because $$\sin\Bigl(\tfrac{\pi}{2}-x\Bigr)=\cos x$$ and $$\cos\Bigl(\tfrac{\pi}{2}-x\Bigr)=\sin x$$, this becomes

$$I = \int_0^{\pi/2} \dfrac{\cos^{3/2}x\,dx}{\cos^{3/2}x+\sin^{3/2}x}$$

Step 2 : Add the two identical expressions for $$I$$

Add the original integral and the transformed one:

$$\begin{aligned} 2I & = \int_0^{\pi/2}\left[\dfrac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x} + \dfrac{\cos^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\right]dx \\[4pt] & = \int_0^{\pi/2} \dfrac{\sin^{3/2}x+\cos^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx \\[4pt] & = \int_0^{\pi/2} 1\,dx \\[4pt] & = \left[x\right]_0^{\pi/2} = \dfrac{\pi}{2}. \end{aligned}$$

Step 3 : Solve for $$I$$

$$2I = \dfrac{\pi}{2}\quad\Rightarrow\quad I = \dfrac{\pi}{4}$$

Hence,

$$\boxed{\displaystyle \int_0^{\pi/2} \dfrac{\sin^{3/2}x\,dx}{\sin^{3/2}x+\cos^{3/2}x} = \dfrac{\pi}{4}}$$

Step 1 : Write the given integral
Let $$I = \int_0^{\pi/2} \dfrac{\cos^5 x}{\sin^5 x + \cos^5 x}\,dx.$$

Step 2 : Apply the property
The definite–integral property
$$\int_0^{a} f(x)\,dx = \int_0^{a} f(a-x)\,dx$$
holds for every integrable function $$f$$. Take
$$a = \dfrac{\pi}{2}, \qquad f(x)=\dfrac{\cos^5 x}{\sin^5 x + \cos^5 x}.$$

Put  $$x = \dfrac{\pi}{2} - t\;(\Rightarrow dx = -dt)$$:

$$\begin{aligned} I &= \int_{\pi/2}^{0} \dfrac{\cos^5 \bigl(\tfrac{\pi}{2}-t\bigr)}{\sin^5 \bigl(\tfrac{\pi}{2}-t\bigr) + \cos^5 \bigl(\tfrac{\pi}{2}-t\bigr)} (-dt) \\[4pt] &= \int_{0}^{\pi/2} \dfrac{\sin^5 t}{\cos^5 t + \sin^5 t}\,dt. \end{aligned}$$

Rename the dummy variable $t$ back to $x$:

$$I = \int_{0}^{\pi/2} \dfrac{\sin^5 x}{\cos^5 x + \sin^5 x}\,dx.$$

Step 3 : Add the two expressions for $$I$$

$$\begin{aligned} 2I &= \int_0^{\pi/2} \left[ \dfrac{\cos^5 x}{\sin^5 x + \cos^5 x} + \dfrac{\sin^5 x}{\cos^5 x + \sin^5 x} \right] dx \\[4pt] &= \int_0^{\pi/2} 1\,dx \\[4pt] &= \left[ x \right]_0^{\pi/2} = \dfrac{\pi}{2}. \end{aligned}$$

Step 4 : Solve for $$I$$
$$I = \dfrac{1}{2}\left(\dfrac{\pi}{2}\right) = \dfrac{\pi}{4}.$$

Hence,
$$\displaystyle \int_0^{\pi/2} \dfrac{\cos^5 x\,dx}{\sin^5 x + \cos^5 x} = \dfrac{\pi}{4}. $$

Given integral: $$I = \int_{-5}^{5}|x+2|\,dx$$

The expression inside the modulus changes sign at $$x=-2$$ (because $$x+2=0$$). Since $$-2$$ lies between the limits $$-5$$ and $$5$$, split the integral at this point:

$$I = \int_{-5}^{-2}|x+2|\,dx + \int_{-2}^{5}|x+2|\,dx$$

Now remove the modulus in each sub-interval.

• For $$x\in[-5,-2]$$, $$x+2<0$$, so $$|x+2| = -(x+2) = -x-2$$.
• For $$x\in[-2,5]$$, $$x+2\ge 0$$, so $$|x+2| = x+2$$.

Hence

$$I = \int_{-5}^{-2}(-x-2)\,dx + \int_{-2}^{5}(x+2)\,dx$$

First part:

$$\int(-x-2)\,dx = -\frac{x^{2}}{2} - 2x$$

Evaluate from $$-5$$ to $$-2$$:

$$\Bigl(-\frac{(-2)^{2}}{2} - 2(-2)\Bigr) - \Bigl(-\frac{(-5)^{2}}{2} - 2(-5)\Bigr) = 2 - \left(-\frac{5}{2}\right) = \frac{9}{2}$$

Second part:

$$\int(x+2)\,dx = \frac{x^{2}}{2} + 2x$$

Evaluate from $$-2$$ to $$5$$:

$$\Bigl(\frac{5^{2}}{2} + 2\cdot5\Bigr) - \Bigl(\frac{(-2)^{2}}{2} + 2(-2)\Bigr) = \frac{45}{2} - (-2) = \frac{49}{2}$$

Total integral:

$$I = \frac{9}{2} + \frac{49}{2} = \frac{58}{2} = 29$$

Thus $$\displaystyle \int_{-5}^{5}|x+2|\,dx = 29$$.

The integrand contains an absolute value. For any real number $$x$$:

• If $$x \lt 5$$: $$|x-5| = 5-x.$$
• If $$x \ge 5$$: $$|x-5| = x-5.$$

The given limits $$2\le x\le 8$$ straddle $$x=5$$, so split the integral there:

$$\int_{2}^{8}|x-5|\,dx = \int_{2}^{5}(5-x)\,dx + \int_{5}^{8}(x-5)\,dx.$$

(i) Integral from 2 to 5.

$$\int_{2}^{5}(5-x)\,dx = \left[5x-\tfrac{x^{2}}{2}\right]_{2}^{5} = \left(25-\tfrac{25}{2}\right) - \left(10-2\right) = \tfrac{25}{2} - 8 = \tfrac{9}{2}.$$

(ii) Integral from 5 to 8.

$$\int_{5}^{8}(x-5)\,dx = \left[\tfrac{x^{2}}{2}-5x\right]_{5}^{8} = \left(32-40\right) - \left(\tfrac{25}{2}-25\right) = -8 - \left(-\tfrac{25}{2}\right) = \tfrac{9}{2}.$$

(iii) Add:

$$\int_{2}^{8}|x-5|\,dx = \tfrac{9}{2}+\tfrac{9}{2} = 9.$$

We have to evaluate the definite integral

$$I = \int_0^1 x\,(1-x)^n\,dx \;\;\;(n \in \mathbb{N}).$$

Step 1 – Use the property $$\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$$.

Take $$f(x)=x(1-x)^n$$ and $$a=0,\;b=1$$. Then

$$I = \int_0^1 \bigl[(1-x)(1-(1-x))^n\bigr]dx = \int_0^1 (1-x)x^n\,dx.$$

Step 2 – Add the two expressions for $$I$$.

$$2I = \int_0^1 \Bigl[x(1-x)^n + (1-x)x^n\Bigr]dx.$$

Step 3 – Split the integral.

$$2I = \int_0^1 x(1-x)^n\,dx + \int_0^1 (1-x)x^n\,dx$$

$$\;\;\; = I + \int_0^1 x^n\,dx - \int_0^1 x^{n+1}\,dx$$

because $$ (1-x)x^n = x^n - x^{n+1}. $$

Step 4 – Isolate $$I$$.

$$2I = I + \int_0^1 x^n\,dx - \int_0^1 x^{n+1}\,dx$$

$$\Rightarrow I = \int_0^1 x^n\,dx - \int_0^1 x^{n+1}\,dx.$$

Step 5 – Evaluate the simple power integrals.

$$\int_0^1 x^n\,dx = \left.\frac{x^{n+1}}{n+1}\right|_0^1 = \frac{1}{n+1},$$

$$\int_0^1 x^{n+1}\,dx = \left.\frac{x^{n+2}}{n+2}\right|_0^1 = \frac{1}{n+2}.$$

Step 6 – Substitute the values.

$$I = \frac{1}{n+1} - \frac{1}{n+2}.$$ Simplify:

$$I = \frac{n+2 - (n+1)}{(n+1)(n+2)} = \frac{1}{(n+1)(n+2)}.$$

Therefore,

$$\boxed{\displaystyle \int_0^1 x(1-x)^n\,dx = \frac{1}{(n+1)(n+2)}}.$$

Let $$I = \int_0^{\pi/4} \log(1+\tan x)\,dx$$ where log denotes the natural logarithm.

Step 1: Apply the property
If $$f(x)$$ is continuous on $$[0,a]$$, then $$\int_0^{a} f(x)\,dx = \int_0^{a} f(a-x)\,dx$$. Taking $$f(x)=\log(1+\tan x)$$ and $$a=\pi/4$$ we have

$$I = \int_0^{\pi/4} \log\bigl(1+\tan(\tfrac{\pi}{4}-x)\bigr)\,dx.$$

Step 2: Simplify the new integrand

• Using $$\tan(\tfrac{\pi}{4}-x)=\dfrac{1-\tan x}{1+\tan x}$$,
• $$1+\tan(\tfrac{\pi}{4}-x)=1+\dfrac{1-\tan x}{1+\tan x}=\dfrac{1+\tan x+1-\tan x}{1+\tan x}=\dfrac{2}{1+\tan x}.$$

Hence

$$I = \int_0^{\pi/4} \log\Bigl(\dfrac{2}{1+\tan x}\Bigr)\,dx = \int_0^{\pi/4}\bigl[\log 2 - \log(1+\tan x)\bigr]dx.$$

Step 3: Separate the integral

$$I = \log 2 \int_0^{\pi/4}dx - \int_0^{\pi/4}\log(1+\tan x)dx = \bigl(\tfrac{\pi}{4}\log 2\bigr) - I.$$

Step 4: Solve for $$I$$

$$I + I = \tfrac{\pi}{4}\log 2 \;\;\Longrightarrow\;\; 2I = \tfrac{\pi}{4}\log 2.$$

$$\therefore\; I = \tfrac{\pi}{8}\log 2.$$

Value of the integral
$$\displaystyle \int_0^{\pi/4}\log(1+\tan x)\,dx = \boxed{\tfrac{\pi}{8}\log 2}. $$

Let $$I = \displaystyle\int_0^2 x\sqrt{2 - x}\,dx.$$

Step 1: Substitute $$u = 2-x$$, so $$x = 2-u$$ and $$dx = -du$$. Limits: $$x=0 \Rightarrow u=2$$ and $$x=2 \Rightarrow u=0$$. Hence

$$I = \int_{2}^{0} (2-u)\sqrt{u}\,(-du) = \int_{0}^{2} (2-u)\sqrt{u}\,du = \int_{0}^{2}\!\left(2u^{1/2} - u^{3/2}\right)du.$$

Step 2: Integrate term by term.

$$I = \left[2\cdot\tfrac{2}{3}u^{3/2} - \tfrac{2}{5}u^{5/2}\right]_{0}^{2} = \left[\tfrac{4}{3}u^{3/2} - \tfrac{2}{5}u^{5/2}\right]_{0}^{2}.$$

Step 3: Evaluate at $$u=2$$. Note $$2^{3/2} = 2\sqrt{2}$$ and $$2^{5/2} = 4\sqrt{2}$$, so

$$I = \tfrac{4}{3}\cdot 2\sqrt{2} - \tfrac{2}{5}\cdot 4\sqrt{2} = \tfrac{8\sqrt{2}}{3} - \tfrac{8\sqrt{2}}{5} = 8\sqrt{2}\!\left(\tfrac{1}{3} - \tfrac{1}{5}\right) = 8\sqrt{2}\cdot\tfrac{2}{15} = \tfrac{16\sqrt{2}}{15}.$$

Let

$$I = \int_{0}^{\pi/2} \bigl(2\,\log \sin x - \log \sin 2x\bigr)\,dx.$$

1. Expand $$\log \sin 2x$$. Since $$\sin 2x = 2\sin x\cos x$$,

$$\log \sin 2x = \log 2 + \log \sin x + \log \cos x.$$

So

$$2\log \sin x - \log \sin 2x = 2\log \sin x - \log 2 - \log \sin x - \log \cos x = \log \sin x - \log \cos x - \log 2.$$

Hence

$$I = \int_{0}^{\pi/2}\log\sin x\,dx - \int_{0}^{\pi/2}\log\cos x\,dx - \int_{0}^{\pi/2}\log 2\,dx.$$

2. Equal sin and cos integrals. With the substitution $$x = \tfrac{\pi}{2}-t$$, the cosine integral becomes the sine integral:

$$\int_{0}^{\pi/2}\log\cos x\,dx = \int_{0}^{\pi/2}\log\sin t\,dt.$$

So the first two terms cancel, leaving

$$I = -\int_{0}^{\pi/2}\log 2\,dx = -\tfrac{\pi}{2}\log 2.$$

Let $$I = \displaystyle\int_{-\pi/2}^{\pi/2} \sin^2 x\,dx.$$

Step 1: Parity. Since $$\sin^2(-x) = \sin^2 x$$, the integrand is even.

Step 2: Use $$\int_{-a}^{a}f(x)\,dx = 2\int_{0}^{a}f(x)\,dx$$.

$$I = 2\int_{0}^{\pi/2}\sin^2 x\,dx.$$

Step 3: Power-reducing identity $$\sin^2 x = \dfrac{1-\cos 2x}{2}$$:

$$I = 2\int_{0}^{\pi/2}\frac{1-\cos 2x}{2}\,dx = \int_{0}^{\pi/2}(1-\cos 2x)\,dx.$$

Step 4: Integrate.

$$I = \left[x - \tfrac{\sin 2x}{2}\right]_{0}^{\pi/2} = \left(\tfrac{\pi}{2} - \tfrac{\sin\pi}{2}\right) - 0 = \tfrac{\pi}{2}.$$

Let

$$I=\int_{0}^{\pi}\dfrac{x\,dx}{1+\sin x}.$$

Step 1: Apply the property $$\displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx$$. Putting $$x \mapsto \pi - x$$ and using $$\sin(\pi-x)=\sin x$$:

$$I = \int_{0}^{\pi}\dfrac{\pi-x}{1+\sin x}\,dx.$$

Step 2: Add the two expressions for $$I$$.

Define $$J = \int_{0}^{\pi}\dfrac{dx}{1+\sin x}$$, so $$I = \tfrac{\pi}{2}J.$$

Step 3: Evaluate $$J$$. Multiply numerator and denominator by $$1-\sin x$$:

$$\dfrac{1}{1+\sin x} = \dfrac{1-\sin x}{1-\sin^{2}x} = \dfrac{1-\sin x}{\cos^{2}x} = \sec^{2}x - \sec x\tan x.$$

The combined antiderivative is

$$\int\!\left(\sec^{2}x - \sec x\tan x\right)dx = \tan x - \sec x + C,$$

which is continuous on $$(0,\pi)$$ except at $$x=\tfrac{\pi}{2}$$, where it has a removable singularity (indeed $$\tan x - \sec x = \dfrac{\sin x - 1}{\cos x} = \dfrac{-\cos^{2}(x/2-\pi/4)\cdot 2}{\ldots}$$; more simply, multiplying numerator and denominator by $$\sin x + 1$$ gives $$\dfrac{-\cos x}{\sin x + 1}$$, which is continuous at $$x=\tfrac{\pi}{2}$$ with value $$0$$). So the antiderivative $$F(x) = -\dfrac{\cos x}{1+\sin x}$$ is continuous on $$[0,\pi]$$, and

$$J = F(\pi) - F(0) = -\dfrac{-1}{1+0} - \left(-\dfrac{1}{1+0}\right) = 1 + 1 = 2.$$

Step 4: Final value of $$I$$.

$$I = \tfrac{\pi}{2} \cdot 2 = \pi.$$

Let $$I = \displaystyle\int_{-\pi/2}^{\pi/2} \sin^7 x\,dx$$.

Consider the integrand $$f(x)=\sin^7 x$$.

Test its symmetry about the origin:

$$f(-x)=\sin^7(-x)=(\! -\sin x)^{7} =-\sin^7 x =-f(x).$$

Thus $$f(x)$$ is an odd function.

For every odd function $$f$$ that is integrable on the symmetric interval $$[-a,a]$$, the definite integral satisfies the property

$$\int_{-a}^{a} f(x)\,dx = 0.$$

Here, the limits are symmetric with $$a=\pi/2$$, so

$$I = \int_{-\pi/2}^{\pi/2} \sin^7 x\,dx = 0.$$

Let $$I = \displaystyle\int_0^{2\pi} \cos^5 x\,dx.$$

We first split the interval $$[0,2\pi]$$ at $$\pi$$:

$$I = \int_{0}^{\pi} \cos^5 x\,dx + \int_{\pi}^{2\pi} \cos^5 x\,dx.$$

Handle the second integral by the substitution

• $$x = \pi + t \;\;(0 \le t \le \pi),$$ so $$dx = dt.$$

Then

$$\int_{\pi}^{2\pi} \cos^5 x\,dx = \int_{0}^{\pi} \cos^5(\pi + t)\,dt.$$

Using the periodic property $$\cos(\pi + t) = -\cos t,$$ and noting that the power $$5$$ is odd, we get

$$\cos^5(\pi + t) = ( -\cos t )^5 = -\cos^5 t.$$

Hence

$$\int_{\pi}^{2\pi} \cos^5 x\,dx = \int_{0}^{\pi} \bigl(-\cos^5 t\bigr)\,dt = -\int_{0}^{\pi} \cos^5 t\,dt.$$

Therefore

$$I = \int_{0}^{\pi} \cos^5 x\,dx - \int_{0}^{\pi} \cos^5 x\,dx = 0.$$

Result: $$\displaystyle\int_0^{2\pi} \cos^5 x\,dx = 0.$$

Let $$I = \int_0^{\pi/2} \dfrac{\sin x - \cos x}{1 + \sin x \cos x}\,dx$$

Using the property of definite integrals

$$\int_0^{a} f(x)\,dx = \int_0^{a} f(a - x)\,dx$$,

we substitute $$a = \dfrac{\pi}{2}$$ and write

$$I = \int_0^{\pi/2} f\!\left(\dfrac{\pi}{2}-x\right)dx,$$ where $$f(x)=\dfrac{\sin x-\cos x}{1+\sin x\cos x}.$$

Evaluate $$f\!\left(\dfrac{\pi}{2}-x\right):$$

Numerator:

$$\sin\!\left(\dfrac{\pi}{2}-x\right)-\cos\!\left(\dfrac{\pi}{2}-x\right)=\cos x-\sin x=-(\sin x-\cos x).$$

Denominator:

$$1+\sin\!\left(\dfrac{\pi}{2}-x\right)\cos\!\left(\dfrac{\pi}{2}-x\right)=1+\cos x\sin x=1+\sin x\cos x.$$

Thus

$$f\!\left(\dfrac{\pi}{2}-x\right)= -\dfrac{\sin x-\cos x}{1+\sin x\cos x}= -f(x).$$

Therefore

$$I = \int_0^{\pi/2} f\!\left(\dfrac{\pi}{2}-x\right)dx = \int_0^{\pi/2} \bigl[-f(x)\bigr]dx = -\int_0^{\pi/2} f(x)dx = -I.$$

Add $$I$$ to both sides:

$$I + I = 0 \quad\Rightarrow\quad 2I = 0 \quad\Rightarrow\quad I = 0.$$

Hence, the required value of the integral is $$0.$$

Let $$I = \displaystyle\int_0^{\pi}\log(1+\cos x)\,dx.$$

Step 1: Half-angle identity $$1+\cos x = 2\cos^{2}\!\tfrac{x}{2}$$:

$$\log(1+\cos x) = \log 2 + 2\log\!\left(\cos\tfrac{x}{2}\right).$$

Hence

$$I = \pi\log 2 + 2\int_{0}^{\pi}\log\!\left(\cos\tfrac{x}{2}\right)dx.$$

Step 2: Substitute $$t = \tfrac{x}{2}$$, so $$dx = 2\,dt$$ and limits become $$0$$ to $$\tfrac{\pi}{2}$$:

$$2\int_{0}^{\pi}\log\!\left(\cos\tfrac{x}{2}\right)dx = 4\int_{0}^{\pi/2}\log\cos t\,dt = 4J,$$

where $$J = \int_{0}^{\pi/2}\log\cos t\,dt.$$

Step 3: Evaluate $$J$$. Using $$\sin\!\left(\tfrac{\pi}{2}-t\right)=\cos t$$:

$$J = \int_{0}^{\pi/2}\log\sin t\,dt.$$

From NCERT Example 34 (or by the standard derivation), $$J = -\tfrac{\pi}{2}\log 2.$$

Step 4: Combine.

$$I = \pi\log 2 + 4\!\left(-\tfrac{\pi}{2}\log 2\right) = \pi\log 2 - 2\pi\log 2 = -\pi\log 2.$$

Let $$I = \int_0^a \dfrac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}}\,dx.$$

Step 1 : Apply the property $$\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx.$$

With $$f(x)=\dfrac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}},$$ we have

$$I = \int_0^a f(a-x)\,dx = \int_0^a \dfrac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}\,dx.$$

Step 2 : Add the two expressions for $$I$$.

$$2I = \int_0^a \left[ f(x) + f(a-x) \right] dx$$
$$\phantom{2I} = \int_0^a \left[ \dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} + \dfrac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}} \right] dx.$$

Since the denominators are identical,

$$f(x)+f(a-x)=\dfrac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}} = 1.$$

Thus

$$2I = \int_0^a 1\,dx = [x]_0^a = a.$$

Step 3 : Solve for $$I$$.

$$2I = a \;\Longrightarrow\; I = \dfrac{a}{2}.$$

Therefore,

$$\displaystyle \int_0^a \dfrac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}}\,dx = \dfrac{a}{2}.$$

Step 1: Locate the sign change. $$|x-1| = 0$$ at $$x=1$$, which lies in $$[0,4]$$. Split the integral there.

Step 2: Rewrite without the modulus.

• For $$0 \le x \le 1$$: $$x-1 \le 0$$, so $$|x-1| = 1-x.$$
• For $$1 \le x \le 4$$: $$x-1 \ge 0$$, so $$|x-1| = x-1.$$

Therefore

$$\int_0^4 |x-1|\,dx = \int_0^1 (1-x)\,dx + \int_1^4 (x-1)\,dx.$$

Step 3: Evaluate each part.

• $$\int_0^1 (1-x)\,dx = \left[x - \tfrac{x^2}{2}\right]_0^1 = 1 - \tfrac12 = \tfrac12.$$
• $$\int_1^4 (x-1)\,dx = \left[\tfrac{x^2}{2}-x\right]_1^4 = (8-4) - (\tfrac12 - 1) = 4 + \tfrac12 = \tfrac92.$$

Step 4: Add.

$$\int_0^4 |x-1|\,dx = \tfrac12 + \tfrac92 = 5.$$

Given relations

• $$f(x)=f(a-x)\;\;\;(0\le x\le a)$$     — so $$f$$ is symmetrical about $$x=\dfrac a2$$.
• $$g(x)+g(a-x)=4\;\;\;(0\le x\le a).$$

Let

$$I=\int_0^a f(x)\,g(x)\,dx.$$

Step 1  :  Re–write the integral by the change of variable

Put $$u=a-x$$. Then $$x=a-u$$ and $$dx=-du$$. When $$x=0$$, $$u=a$$ and when $$x=a$$, $$u=0$$. Therefore

$$I=\int_{x=0}^{x=a}f(x)\,g(x)\,dx=\int_{u=a}^{u=0}f(a-u)\,g(a-u)\,(-du)=\int_{0}^{a}f(a-u)\,g(a-u)\,du.$$

Since the dummy variable can again be called $$x$$, this becomes

$$I=\int_0^a f(a-x)\,g(a-x)\,dx.$$

Step 2  :  Employ the symmetry of $$f$$

Because $$f(a-x)=f(x)$$ for every $$x$$ in $$[0,a]$$, the last integral is

$$I=\int_0^a f(x)\,g(a-x)\,dx.$$

Step 3  :  Add the two expressions for $$I$$

We now possess two expressions:

• (i) $$I=\int_0^a f(x)\,g(x)\,dx$$
• (ii) $$I=\int_0^a f(x)\,g(a-x)\,dx$$

Add them term by term:

$$2I=\int_0^a\bigl[f(x)\,g(x)+f(x)\,g(a-x)\bigr]dx=\int_0^a f(x)\,[g(x)+g(a-x)]\,dx.$$

Step 4  :  Insert the relation for $$g$$

Given $$g(x)+g(a-x)=4$$, hence

$$2I=\int_0^a f(x)\,(4)\,dx=4\int_0^a f(x)\,dx.$$

Step 5  :  Solve for $$I$$

Divide by $$2$$:

$$I=2\int_0^a f(x)\,dx.$$

That is,

$$\int_0^a f(x)\,g(x)\,dx = 2\int_0^a f(x)\,dx,$$

which is exactly what we had to prove.

Given integral

$$I=\displaystyle\int_{-\pi/2}^{\pi/2}\,(x^{3}+x\cos x+\tan^{5}x+1)\,dx$$

Over a symmetric interval $$[-a,a]$$ the following facts are very useful:

• If $$f(x)$$ is even, i.e. $$f(-x)=f(x)$$, then $$\displaystyle\int_{-a}^{a}f(x)\,dx=2\int_{0}^{a}f(x)\,dx.$$
• If $$g(x)$$ is odd, i.e. $$g(-x)=-g(x)$$ and the integral exists, then $$\displaystyle\int_{-a}^{a}g(x)\,dx=0.$$

Therefore we first examine each term in the integrand.

1. $$x^{3}$$

$$(-x)^{3}=-x^{3}\;\Rightarrow\;x^{3}$$ is odd.

2. $$x\cos x$$

Since $$x$$ is odd and $$\cos x$$ is even, their product is odd:

$$x\cos x\;\text{is odd.}$$

3. $$\tan^{5}x$$

Because $$\tan(-x)=-\tan x$$ (odd) and the power 5 is also odd,

$$\tan^{5}(-x)=-(\tan^{5}x)\;\Rightarrow\;\tan^{5}x\;\text{is odd.}$$

4. Constant 1

The constant function satisfies $$1=1$$ for every $$x$$, hence it is even.

Split the integral

$$I=\int_{-\pi/2}^{\pi/2}x^{3}\,dx+\int_{-\pi/2}^{\pi/2}x\cos x\,dx+\int_{-\pi/2}^{\pi/2}\tan^{5}x\,dx+\int_{-\pi/2}^{\pi/2}1\,dx.$$

Using parity:

• The first three integrands are odd, so their integrals over $$[-\pi/2,\,\pi/2]$$ are zero.
• The last integrand is even, so we have

$$\int_{-\pi/2}^{\pi/2}1\,dx=\bigl(x\bigr)\Big|_{-\pi/2}^{\pi/2}=\pi/2-(-\pi/2)=\pi.$$

Hence

$$I=0+0+0+\pi=\pi.$$

The correct alternative is (C) $$\pi$$.

Let
$$I = \displaystyle\int_0^{\pi/2} \log\left(\dfrac{4 + 3\sin x}{4 + 3\cos x}\right)\,dx$$

Split the logarithm into two simpler integrals:

$$I = \int_0^{\pi/2} \log(4 + 3\sin x)\,dx \; - \; \int_0^{\pi/2} \log(4 + 3\cos x)\,dx.$$

Denote

$$A = \int_0^{\pi/2} \log(4 + 3\sin x)\,dx, \qquad B = \int_0^{\pi/2} \log(4 + 3\cos x)\,dx.$$

Thus $$I = A - B.$$

Show that $$A = B.$$

• Start from $$A = \displaystyle \int_0^{\pi/2} \log(4 + 3\sin x)\,dx.$$
• Use the substitution $$x = \dfrac{\pi}{2} - t\;\;(\Rightarrow dx = -dt).$$
• When $$x = 0,\; t = \dfrac{\pi}{2}$$; when $$x = \dfrac{\pi}{2},\; t = 0.$$
• Therefore

$$A = \int_{\pi/2}^{0} \log\bigl(4 + 3\sin(\tfrac{\pi}{2}-t)\bigr)(-dt) = \int_0^{\pi/2} \log\bigl(4 + 3\cos t\bigr)\,dt.$$

But the right–hand integral is exactly $$B.$$ Hence $$A = B.$$

Conclude the value of $$I.$$ Since $$I = A - B$$ and $$A = B,$$ we get

$$I = 0.$$

The correct option is (C) 0.

The integral is $$I = \displaystyle\int \cos 6x\sqrt{1+\sin 6x}\,dx$$.

Let $$u = 1+\sin 6x$$. Then

$$du = 6\cos 6x\,dx \;\;\Longrightarrow\;\; \cos 6x\,dx = \dfrac{du}{6}.$$

Substitute in the integral:

$$I = \int \sqrt{u}\;\dfrac{du}{6} = \dfrac{1}{6}\int u^{1/2}\,du.$$

Using the power rule,

$$\int u^{1/2}\,du = \dfrac{u^{3/2}}{3/2} = \dfrac{2}{3}u^{3/2}.$$

Therefore

$$I = \dfrac{1}{6}\cdot \dfrac{2}{3}u^{3/2} + C = \dfrac{1}{9}u^{3/2}+C.$$

Re-substitute $$u = 1+\sin 6x$$:

$$I = \dfrac{1}{9}\left(1+\sin 6x\right)^{3/2}+C.$$

Hence,

$$\displaystyle\int \cos 6x\sqrt{1+\sin 6x}\,dx = \dfrac{1}{9}(1+\sin 6x)^{3/2}+C.$$

Let $$I = \displaystyle \int \frac{(x^{4}-x)^{1/4}}{x^{5}}\,dx$$.

1. First substitution  $$x = \dfrac1t$$

Then $$t = \dfrac1x\;,\; dx = -\dfrac{1}{t^{2}}\,dt$$.

Evaluate the expression inside the integral:

$$x^{4}-x = \frac1{t^{4}}-\frac1t = \frac{1-t^{3}}{t^{4}} \;\Longrightarrow\; (x^{4}-x)^{1/4}=\Bigl(\frac{1-t^{3}}{t^{4}}\Bigr)^{1/4}= \frac{(1-t^{3})^{1/4}}{t}\;.$$

Also $$x^{5}=\frac1{t^{5}}\;.$$

Substituting in $$I$$ gives

$$I = \int \frac{(1-t^{3})^{1/4}/t}{1/t^{5}} \cdot \bigl(-\frac1{t^{2}}\,dt\bigr)= -\int t^{2}(1-t^{3})^{1/4}\,dt.$$

2. Second substitution  $$u = 1-t^{3}$$

Then $$du = -3t^{2}\,dt \;\Longrightarrow\; t^{2}\,dt = -\frac{du}{3}$$.

Therefore

$$I = -\int t^{2}(1-t^{3})^{1/4}\,dt = -\int (1-t^{3})^{1/4}\,t^{2}\,dt = -\int u^{1/4}\Bigl(-\frac{du}{3}\Bigr)= \frac13\int u^{1/4}\,du.$$

3. Integrate

$$\int u^{1/4}\,du = \frac{u^{5/4}}{5/4}=\frac{4}{5}u^{5/4}.$$

Hence $$I = \frac13\cdot\frac45u^{5/4}+C = \frac{4}{15}u^{5/4}+C.$$

4. Back–substitute

Recall $$u = 1-t^{3},\; t = \frac1x \;\Rightarrow\; t^{3}=\frac1{x^{3}}.$$

Thus $$u = 1-\frac1{x^{3}} = \frac{x^{3}-1}{x^{3}}$$ and

$$u^{5/4}=\Bigl(\frac{x^{3}-1}{x^{3}}\Bigr)^{5/4}=\frac{(x^{3}-1)^{5/4}}{x^{15/4}}.$$

Finally

$$I = \frac{4}{15}\,\frac{(x^{3}-1)^{5/4}}{x^{15/4}} + C.$$

We have to evaluate the integral

$$I=\displaystyle\int \dfrac{x^4\,dx}{(x-1)(x^2+1)}.$$

1. Reduce the degree by polynomial long-division

The denominator is

$$D(x)=(x-1)(x^2+1)=x^3-x^2+x-1.$$

Divide the numerator by this cubic:

$$\begin{aligned} x^4 &= (x)\,D(x)+\bigl(x^3-x^2+x\bigr)\\[4pt] &= (x+1)\,D(x)+1. \end{aligned}$$

Hence

$$\dfrac{x^4}{D(x)}=x+1+\dfrac{1}{D(x)}$$

and therefore

$$I=\int (x+1)\,dx+\int\dfrac{dx}{(x-1)(x^2+1)}.$$

2. Integrate the polynomial part

$$\int(x+1)\,dx=\dfrac{x^2}{2}+x.$$

3. Partial-fraction decomposition of $$\dfrac{1}{(x-1)(x^2+1)}$$

Assume

$$\dfrac{1}{(x-1)(x^2+1)}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+1}.$$

Clearing denominators gives

$$1=A(x^2+1)+(Bx+C)(x-1).$$

Expanding and collecting like powers of $$x$$ produces

$$1=(A+B)x^2+(-B+C)x+(A-C).$$

Equating coefficients to those of the constant polynomial $$1$$ yields

Solving the system:

$$B=-A,\qquad C=-A,\qquad 2A=1\;\Longrightarrow\;A=\dfrac12,$$

so that

$$B=C=-\dfrac12.$$

Thus

$$\dfrac{1}{(x-1)(x^2+1)}=\dfrac{1/2}{x-1}-\dfrac{1}{2}\,\dfrac{x+1}{x^2+1}.$$

4. Integrate the partial fractions

Write

$$J=\int\dfrac{dx}{(x-1)(x^2+1)}=\dfrac12\int\dfrac{dx}{x-1}-\dfrac12\int\dfrac{x+1}{x^2+1}\,dx.$$

First term:

$$\dfrac12\int\dfrac{dx}{x-1}=\dfrac12\ln|x-1|.$$

Second term – split the numerator:

$$\int\dfrac{x+1}{x^2+1}\,dx=\int\dfrac{x}{x^2+1}\,dx+\int\dfrac{dx}{x^2+1}.$$

• For $$\int \dfrac{x}{x^2+1}\,dx$$ put $$u=x^2+1$$, $$du=2x\,dx\;\Rightarrow\;dx=du/(2x).$$ Directly,

$$\int\dfrac{x}{x^2+1}\,dx=\dfrac12\ln(x^2+1).$$

• The remaining integral is the standard arctangent:

$$\int\dfrac{dx}{x^2+1}=\tan^{-1}x.$$

Putting these together,

$$\int\dfrac{x+1}{x^2+1}\,dx=\dfrac12\ln(x^2+1)+\tan^{-1}x.$$

Therefore

$$J=\dfrac12\ln|x-1|-\dfrac12\Bigl[\dfrac12\ln(x^2+1)+\tan^{-1}x\Bigr] =\dfrac12\ln|x-1|-\dfrac14\ln(x^2+1)-\dfrac12\tan^{-1}x.$$

5. Combine all parts

Adding the result of step 2 and step 4, we finally get

$$I=\dfrac{x^2}{2}+x+\dfrac12\ln|x-1|-\dfrac14\ln(x^2+1)-\dfrac12\tan^{-1}x+C,$$

where $$C$$ is the constant of integration.

We have to evaluate

$$I = \displaystyle \int \left[\log(\log x) + \dfrac{1}{(\log x)^2}\right]\,dx.$$

Throughout the calculation the ordinary logarithm is the natural logarithm (base e).

Step 1 : Substitute $$t = \log x.$$

Then $$dt = \dfrac{dx}{x}\;\Longrightarrow\;dx = x\,dt.$$
Since $$x = e^t,$$ we get $$dx = e^{t}\,dt.$$

Step 2 : Rewrite the integral in the new variable t.

\[\begin{aligned} I &= \int \Bigl[\log(\log x) + \dfrac{1}{(\log x)^2}\Bigr]dx\\ &= \int \Bigl[\log t + \dfrac{1}{t^{2}}\Bigr] \, e^{t}\,dt\\ &= \int e^{t}\,\log t\,dt \; + \; \int \dfrac{e^{t}}{t^{2}}\,dt.\tag{1} \end{aligned}\]

Step 3 : Combine the two integrals cleverly.

Observe the derivative of the expression $$e^{t}\,\log t - \dfrac{e^{t}}{t}$$ :

$$\begin{aligned} \dfrac{d}{dt}\Bigl(e^{t}\,\log t - \dfrac{e^{t}}{t}\Bigr) &= \bigl(e^{t}\,\log t + e^{t}\,\dfrac{1}{t}\bigr) - \Bigl(\dfrac{e^{t}}{t} - \dfrac{e^{t}}{t^{2}}\Bigr)\\[4pt] &= e^{t}\,\log t \; + \; \cancel{e^{t}/t} - \cancel{e^{t}/t} \; + \; \dfrac{e^{t}}{t^{2}}\\[4pt] &= e^{t}\,\log t + \dfrac{e^{t}}{t^{2}}. \end{aligned}$$

This derivative is exactly the integrand in (1). Hence

$$I = e^{t}\,\log t - \dfrac{e^{t}}{t} + C,$$ where $$C$$ is the constant of integration.

Step 4 : Return to the original variable x.

Because $$e^{t}=x$$ and $$t=\log x,$$ we finally obtain

$$I = x\,\log(\log x) - \dfrac{x}{\log x} + C.$$

Let $$I = \displaystyle\int \bigl[\sqrt{\cot x}+\sqrt{\tan x}\bigr]\,dx.$$ Throughout the region in which both square roots are real we have $$\sin x > 0$$ and $$\cos x > 0$$ (otherwise the integrand is not defined).

Step 1: Combine the radicals.

$$\sqrt{\tan x}+\sqrt{\cot x}=\dfrac{\sin x}{\sqrt{\sin x\cos x}}+\dfrac{\cos x}{\sqrt{\sin x\cos x}}=\dfrac{\sin x+\cos x}{\sqrt{\sin x\cos x}}.$$

Multiply numerator and denominator by $$\sqrt{2}$$:

$$I = \sqrt2 \int \dfrac{\sin x + \cos x}{\sqrt{2\sin x\cos x}}\,dx = \sqrt2\int\dfrac{\sin x+\cos x}{\sqrt{\sin 2x}}\,dx.$$

Step 2: Substitute $$t = \sin x - \cos x.$$ Then $$dt = (\cos x + \sin x)\,dx$$ and

$$t^{2} = (\sin x - \cos x)^{2} = 1 - 2\sin x\cos x = 1 - \sin 2x,$$

so $$\sin 2x = 1 - t^{2}.$$ Substituting:

$$I = \sqrt2 \int \dfrac{dt}{\sqrt{1-t^{2}}}.$$

Step 3: Evaluate. Using $$\int \dfrac{dt}{\sqrt{1-t^{2}}}=\sin^{-1}t$$:

$$I = \sqrt2\,\sin^{-1}t + C = \sqrt2\,\sin^{-1}\!\bigl(\sin x - \cos x\bigr) + C.$$

We wish to evaluate $$\displaystyle\int \dfrac{\sin 2x\,\cos 2x\,dx}{\sqrt{9-\cos^{4}(2x)}}$$.

Step 1: Substitute an expression whose derivative already appears.
Let $$t = \cos^{2} 2x$$. Then

$$\dfrac{dt}{dx} = \dfrac{d}{dx}\,(\cos^{2} 2x) = 2\cos 2x\,( -\sin 2x)\,\cdot 2 = -4\sin 2x\,\cos 2x.$$

Hence $$\sin 2x\,\cos 2x\,dx = -\dfrac{1}{4}\,dt.$$

Step 2: Rewrite the integral in terms of t.
Because $$\cos^{4} 2x = (\cos^{2} 2x)^{2} = t^{2},$$ we obtain

$$\int \dfrac{\sin 2x\,\cos 2x\,dx}{\sqrt{9-\cos^{4} 2x}} = \int \dfrac{-\dfrac14\,dt}{\sqrt{9-t^{2}}} = -\dfrac14\int \dfrac{dt}{\sqrt{9-t^{2}}}.$$

Step 3: Integrate the standard form.
For any constant $$a>0$$, $$\int \dfrac{dt}{\sqrt{a^{2}-t^{2}}}=\sin^{-1}\!\left(\dfrac{t}{a}\right)+C.$$ Here $$a=3.$$ Therefore

$$-\dfrac14\int \dfrac{dt}{\sqrt{9-t^{2}}} = -\dfrac14\,\sin^{-1}\!\left(\dfrac{t}{3}\right)+C.$$

Step 4: Back-substitute $$t=\cos^{2}2x$$.

$$\displaystyle\int \dfrac{\sin 2x\,\cos 2x\,dx}{\sqrt{9-\cos^{4}(2x)}} = -\dfrac14\,\sin^{-1}\!\left(\dfrac{\cos^{2}2x}{3}\right)+C.$$

Step 1 : Locate the points where the integrand changes sign

The quantity inside the modulus is $$f(x)=x\sin(\pi x)$$. It can be zero only when

• $$x=0$$ (because of the factor $$x$$), or
• $$\sin(\pi x)=0\;\Longrightarrow\;x\in\mathbb Z$$.

Inside the interval $$[-1,\;3/2]$$ the integral zeros are therefore at $$x=-1,\;0,\;1$$.

Step 2 : Determine the sign of $$f(x)$$ on each sub-interval

Hence

$$|x\sin(\pi x)|=\begin{cases} x\sin(\pi x), & -1\le x\le 1,\\ -\,x\sin(\pi x), & 1\le x\le 3/2. \end{cases}$$

Step 3 : Split the definite integral

$$\displaystyle I=\int_{-1}^{3/2}|x\sin(\pi x)|\,dx =\int_{-1}^{0}x\sin(\pi x)\,dx+ \int_{0}^{1}x\sin(\pi x)\,dx- \int_{1}^{3/2}x\sin(\pi x)\,dx.$$

Step 4 : Find an antiderivative of $$x\sin(\pi x)$$

Use integration by parts with $$u=x,\;dv=\sin(\pi x)\,dx$$:

$$v=\int\sin(\pi x)\,dx=-\,\frac{\cos(\pi x)}{\pi},\qquad du=dx.$$

\[\begin{aligned} \int x\sin(\pi x)\,dx&=x\,v-\int v\,du\\[2pt] &=-x\frac{\cos(\pi x)}{\pi}+\frac1\pi\int\cos(\pi x)\,dx\\[2pt] &=-x\frac{\cos(\pi x)}{\pi}+\frac{\sin(\pi x)}{\pi^{2}}+C. \end{aligned}\]

Define $$F(x)= -\,\dfrac{x\cos(\pi x)}{\pi}+\dfrac{\sin(\pi x)}{\pi^{2}}.$$ Then $$F'(x)=x\sin(\pi x).$$

Step 5 : Evaluate each part

(a) Interval [−1,0]

$$I_{1}=F(0)-F(-1).$$ $$F(0)=0,\qquad F(-1)=-\,\frac{1}{\pi}.$$ Thus $$I_{1}=0-\Bigl(-\frac1\pi\Bigr)=\frac1\pi.$$

(b) Interval [0,1]

$$I_{2}=F(1)-F(0).$$ $$F(1)=\frac1\pi,\qquad F(0)=0.$$ Hence $$I_{2}=\frac1\pi.$$

(c) Interval [1,3/2]

$$I_{3}=-\bigl[F(3/2)-F(1)\bigr].$$ $$F\!\left(\tfrac32\right)=-\,\frac1{\pi^{2}},\qquad F(1)=\frac1\pi.$$ Therefore

$$I_{3}=-\left(-\frac1{\pi^{2}}-\frac1\pi\right)=\frac1{\pi^{2}}+\frac1\pi.$$

Step 6 : Add the three contributions

\[\begin{aligned} I&=I_{1}+I_{2}+I_{3}=\frac1\pi+\frac1\pi+\left(\frac1\pi+\frac1{\pi^{2}}\right)\\[4pt] &=\frac{3}{\pi}+\frac{1}{\pi^{2}}=\frac{3\pi+1}{\pi^{2}}. \end{aligned}\]

Step 7 : Final result

The required integral equals $$\displaystyle\boxed{\dfrac{3\pi+1}{\pi^{2}}}$$.

Let
$$I = \int_{0}^{\pi}\dfrac{x\,dx}{a^{2}\cos^{2}x + b^{2}\sin^{2}x}.$$

Step 1 – Apply the property $$\int_{0}^{A}f(x)\,dx = \int_{0}^{A}f(A-x)\,dx$$.
Putting $$x \mapsto \pi - x$$ gives $$I = \int_{0}^{\pi}\dfrac{\pi - x}{a^{2}\cos^{2}x + b^{2}\sin^{2}x}\,dx.$$ Adding the two expressions of $$I$$ we get $$2I = \int_{0}^{\pi}\dfrac{x + (\pi - x)}{a^{2}\cos^{2}x + b^{2}\sin^{2}x}\,dx = \pi\int_{0}^{\pi}\dfrac{dx}{a^{2}\cos^{2}x + b^{2}\sin^{2}x}.$$

Hence
$$I = \dfrac{\pi}{2}\,J,$$ where $$J = \int_{0}^{\pi}\dfrac{dx}{a^{2}\cos^{2}x + b^{2}\sin^{2}x}.$$

Step 2 – Use symmetry of the integrand.
Because $$\cos^{2}(\pi - x)=\cos^{2}x$$ and $$\sin^{2}(\pi - x)=\sin^{2}x$$, the integrand is symmetric about $$x = \dfrac{\pi}{2}$$. Consequently $$J = 2\int_{0}^{\frac{\pi}{2}}\dfrac{dx}{a^{2}\cos^{2}x + b^{2}\sin^{2}x}.$$

Step 3 – Convert to an improper rational integral.
Put $$t = \tan x\;(0 \le x \le \tfrac{\pi}{2})$$ so that $$dx = \dfrac{dt}{1+t^{2}}$$, $$\cos^{2}x = \dfrac{1}{1+t^{2}},\quad \sin^{2}x = \dfrac{t^{2}}{1+t^{2}}.$$
Then $$a^{2}\cos^{2}x + b^{2}\sin^{2}x = \dfrac{a^{2} + b^{2}t^{2}}{1+t^{2}},$$ so \[\dfrac{dx}{a^{2}\cos^{2}x + b^{2}\sin^{2}x} = \dfrac{1+t^{2}}{a^{2}+b^{2}t^{2}}\,\dfrac{dt}{1+t^{2}} = \dfrac{dt}{a^{2}+b^{2}t^{2}}.\] The limits become $$t=0$$ when $$x=0$$ and $$t \to \infty$$ when $$x \to \tfrac{\pi}{2}$$. Therefore $$\int_{0}^{\frac{\pi}{2}}\dfrac{dx}{a^{2}\cos^{2}x + b^{2}\sin^{2}x} = \int_{0}^{\infty}\dfrac{dt}{a^{2}+b^{2}t^{2}}.$$

Step 4 – Evaluate the standard integral.
Factorising, $$\int_{0}^{\infty}\dfrac{dt}{a^{2}+b^{2}t^{2}} = \dfrac{1}{a^{2}}\int_{0}^{\infty}\dfrac{dt}{1+\left(\dfrac{b}{a}t\right)^{2}}.$$
Put $$u = \dfrac{b}{a}t \;\Rightarrow\; dt = \dfrac{a}{b}du$$; then \[\int_{0}^{\infty}\dfrac{dt}{a^{2}+b^{2}t^{2}} = \dfrac{1}{a^{2}}\cdot\dfrac{a}{b}\int_{0}^{\infty}\dfrac{du}{1+u^{2}} = \dfrac{1}{ab}\left[\arctan u\right]_{0}^{\infty} = \dfrac{1}{ab}\,\dfrac{\pi}{2} = \dfrac{\pi}{2ab}.\]

Step 5 – Assemble the result.
Thus $$J = 2\times \dfrac{\pi}{2ab} = \dfrac{\pi}{ab},$$ and finally $$I = \dfrac{\pi}{2}\,J = \dfrac{\pi}{2}\times \dfrac{\pi}{ab} = \dfrac{\pi^{2}}{2ab}. $$

We have to evaluate

$$I=\int\dfrac{1}{x-x^{3}}\,dx.$$

1. Factorise the denominator.

$$x-x^{3}=x\,(1-x^{2})=x\,(1-x)(1+x).$$

2. Partial-fraction decomposition. Write

$$\dfrac{1}{x(1-x)(1+x)}=\dfrac{A}{x}+\dfrac{B}{1-x}+\dfrac{C}{1+x}.$$

Multiplying both sides by $$x(1-x)(1+x)$$ gives the identity

\[1=A\,(1-x)(1+x)+B\,x\,(1+x)+C\,x\,(1-x).\tag{*}\]

Now substitute the three convenient values of $$x$$ (the roots of the factors), one at a time:

• Put $$x=0$$ in (*): the $$B$$- and $$C$$-terms vanish, leaving $$1=A\,(1)(1)=A,$$ so $$A=1.$$
• Put $$x=1$$ in (*): the $$A$$- and $$C$$-terms vanish, leaving $$1=B\,(1)(2)=2B,$$ so $$B=\tfrac12.$$
• Put $$x=-1$$ in (*): the $$A$$- and $$B$$-terms vanish, leaving $$1=C\,(-1)(1-(-1))=C\,(-1)(2)=-2C,$$ so $$C=-\tfrac12.$$

Hence

$$\dfrac{1}{x-x^{3}}=\dfrac{1}{x}+\dfrac{1/2}{1-x}-\dfrac{1/2}{1+x}.$$

3. Integrate term by term. Using $$\displaystyle\int\dfrac{dx}{x}=\ln|x|$$ and $$\displaystyle\int\dfrac{dx}{1\mp x}=\mp\ln|1\mp x|$$,

$$\int\dfrac{dx}{x-x^{3}}=\ln|x|-\tfrac12\ln|1-x|-\tfrac12\ln|1+x|+C.$$

Combining the last two logarithms,

$$=\ln|x|-\tfrac12\ln\bigl|(1-x)(1+x)\bigr|+C=\ln|x|-\tfrac12\ln|1-x^{2}|+C.$$

Given integral

Let $$I = \int \frac{1}{\sqrt{x+a}+\sqrt{x+b}}\,dx$$ where $$a,b$$ are real constants and, unless stated otherwise, $$a\neq b$$ (the case $$a=b$$ is handled at the end).

Step 1. Rationalise the denominator

Multiply the integrand by the conjugate $$\sqrt{x+a}-\sqrt{x+b}$$ (numerator and denominator) :

$$I = \int \frac{\sqrt{x+a}-\sqrt{x+b}}{(\sqrt{x+a}+\sqrt{x+b})(\sqrt{x+a}-\sqrt{x+b})}\,dx$$

The denominator is now a difference of squares:

$$ (\sqrt{x+a})^2-(\sqrt{x+b})^2 = (x+a)-(x+b) = a-b. $$

Hence

$$I = \int \frac{\sqrt{x+a}-\sqrt{x+b}}{a-b}\,dx.$$

Step 2. Split the integral

Because $$a-b$$ is constant we may take its reciprocal outside:

$$I = \frac{1}{a-b}\int \bigl(\sqrt{x+a}-\sqrt{x+b}\bigr)\,dx$$

$$\phantom{I}=\frac{1}{a-b}\Bigl[\int\sqrt{x+a}\,dx-\int\sqrt{x+b}\,dx\Bigr].$$

Step 3. Integrate each term separately

Recall $$\int \sqrt{t}\,dt = \tfrac{2}{3}t^{3/2}+C$$. Substitute $$t=x+a$$ in the first integral and $$t=x+b$$ in the second:

$$\int\sqrt{x+a}\,dx = \frac{2}{3}(x+a)^{3/2}+C_1,$$

$$\int\sqrt{x+b}\,dx = \frac{2}{3}(x+b)^{3/2}+C_2.$$

(The two arbitrary constants merge into a single constant in the final answer.)

Step 4. Assemble the result

$$I = \frac{1}{a-b}\left[\frac{2}{3}(x+a)^{3/2}-\frac{2}{3}(x+b)^{3/2}\right]+C$$

or, more compactly,