Integration by Substitution

1. Simplify the Integrand
To evaluate the integral, divide both the numerator and the denominator by $$x^2$$:
$$\int \frac{x^2+1}{x^4+1} \, dx = \int \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} \, dx$$

2. Rewrite the Denominator
We can express the denominator $$x^2 + \frac{1}{x^2}$$ in terms of the algebraic identity $$(x - \frac{1}{x})^2$$:
$$\left(x - \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} - 2 \implies x^2 + \frac{1}{x^2} = \left(x - \frac{1}{x}\right)^2 + 2$$

Substitute this back into our integral expression:
$$\int \frac{1 + \frac{1}{x^2}}{\left(x - \frac{1}{x}\right)^2 + 2} \, dx$$

3. Use Substitution Method
Let us use the substitution method by setting:
$$u = x - \frac{1}{x}$$

Differentiating both sides with respect to $$x$$ gives:
$$du = \left(1 + \frac{1}{x^2}\right) dx$$

Substitute $$u$$ and $$du$$ into the rewritten integral:
$$\int \frac{1}{u^2 + 2} \, du = \int \frac{1}{u^2 + (\sqrt{2})^2} \, du$$

4. Evaluate the Integral
Using the standard integration formula $$\int \frac{1}{y^2 + a^2} \, dy = \frac{1}{a} \tan^{-1}\left(\frac{y}{a}\right) + C$$, we get:
$$\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{u}{\sqrt{2}}\right) + C$$

5. Substitute Back the Original Variable
Replace $$u$$ with its original definition $$x - \frac{1}{x}$$:
$$\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x - \frac{1}{x}}{\sqrt{2}}\right) + C = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x^2 - 1}{\sqrt{2}x}\right) + C$$

To evaluate this symmetric integration problem , we divide the numerator and denominator by powers of $$x$$ to transform the expression into a standard functional form.

Step 1: Divide the numerator and denominator by $$x^2$$,we factor out $$x$$ from the first factor of the denominator, and $$x^2$$ from inside the radical of the second factor then we get:

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

Step 2: Express the variables using substitution

Let $$t = x + \frac{1}{x}$$. Differentiating both sides with respect to $$x$$ gives:

$$dt = \left(1 - \frac{1}{x^2}\right) dx$$

We also need to express the terms inside the radical in terms of $$t$$ by squaring our substitution variable:

$$t^2 = \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2} \implies x^2 + \frac{1}{x^2} = t^2 - 2$$

Substituting $$t = x + \frac{1}{x}$$ directly converts the expression into:

$$I = \int \frac{dt}{t\sqrt{t^2 - 2}}$$

Step 3: Evaluate the standard integral form

To evaluate $$\int \frac{dt}{t\sqrt{t^2 - 2}}$$, we can substitute $$t = \sqrt{2}\sec\theta$$, which gives:

$$dt = \sqrt{2}\sec\theta\tan\theta \, d\theta$$

Substitute these trigonometric terms into the integral expression:

$$I = \int \frac{\sqrt{2}\sec\theta\tan\theta \, d\theta}{\sqrt{2}\sec\theta\sqrt{2\sec^2\theta - 2}}$$

$$I = \int \frac{\sqrt{2}\sec\theta\tan\theta \, d\theta}{\sqrt{2}\sec\theta \cdot \sqrt{2}\tan\theta} = \int \frac{1}{\sqrt{2}} \, d\theta$$

Integrating a constant value gives:

$$I = \frac{1}{\sqrt{2}}\theta + C$$

Step 4: Substitute back to match the target form

Now, replace the temporary variable $$\theta$$ with our initial variable track expressions:

$$\sec\theta = \frac{t}{\sqrt{2}} \implies \theta = \sec^{-1}\left(\frac{t}{\sqrt{2}}\right)$$

Substitute $$t = x + \frac{1}{x} = \frac{x^2 + 1}{x}$$ back into the angle slot:

$$I = \frac{1}{\sqrt{2}}\sec^{-1}\left(\frac{x^2 + 1}{x\sqrt{2}}\right) + C$$

Comparing this directly with the given template in the question $$k \sec^{-1}\left(\frac{x^2 + 1}{x\sqrt{2}}\right) + C$$:

$$k= \frac{1}{\sqrt{2}}$$

First simplify the logarithmic part.

$$\ln(x^2+1)-2\ln x=\ln\left(\frac{x^2+1}{x^2}\right)$$

$$=\ln\left(1+\frac1{x^2}\right)$$

Therefore,

$$I=\int \frac{\sqrt{x^2+1}}{x^4}\ln\left(1+\frac1{x^2}\right)\,dx$$

Notice that

$$\frac{\sqrt{x^2+1}}{x^4}=\frac1{x}\left(1+\frac1{x^2}\right)^{1/2}\frac1{x^2}$$

Let $$t=\sqrt{1+\frac1{x^2}}$$

Then $$t^2=1+\frac1{x^2}$$

Differentiating,

$$2t\,dt=-\frac{2}{x^3}\,dx$$

$$dt=-\frac1{x^3t}\,dx$$

$$dx=-x^3t\,dt$$

Also, $$\ln\left(1+\frac1{x^2}\right)=\ln(t^2)=2\ln t$$

Substituting in the integral,

$$I=\int \frac{t}{x}\cdot\frac1{x^2}\cdot(2\ln t)\cdot(-x^3t)\,dt$$

$$=-2\int t^2\ln t\,dt$$

Now integrate by parts.

Let $$u=\ln t,\qquad dv=t^2dt$$

Then $$du=\frac{dt}{t},\qquad v=\frac{t^3}{3}$$

Therefore, $$\int t^2\ln t\,dt=\frac{t^3}{3}\ln t-\frac13\int t^2dt$$

$$=\frac{t^3}{3}\ln t-\frac{t^3}{9}$$

Hence,

$$I=-2\left(\frac{t^3}{3}\ln t-\frac{t^3}{9}\right)+C$$

$$=-\frac23t^3\ln t+\frac29t^3+C$$

Factor out

$$\frac13t^3$$

$$I=\frac13t^3\left(\frac23-2\ln t\right)+C$$

Using

$$2\ln t=\ln(t^2)$$

and

$$t^2=1+\frac1{x^2}$$

we get

$$I=\frac13\left(1+\frac1{x^2}\right)^{3/2}\left[\frac23-\ln\left(1+\frac1{x^2}\right)\right]+C$$

Final Answer :

$$\frac13\left(1+\frac1{x^2}\right)^{3/2}\left[\frac23-\ln\left(1+\frac1{x^2}\right)\right]+C$$