Constant of Integration

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

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 :

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

We need to evaluate $$\displaystyle I = \int \frac{\ln x}{x^{2}}\,dx$$ for $$x \gt 0$$.

Step 1: Substitution
Put $$t = \ln x \;\; \Longrightarrow \;\; x = e^{t}, \qquad dx = e^{t}\,dt = x\,dt.$$

Rewrite the integrand using this substitution:
$$I = \int \frac{\ln x}{x^{2}}\,dx = \int \frac{t}{x^{2}}\,(x\,dt) = \int \frac{t}{x}\,dt = \int t\,e^{-t}\,dt,$$ because $$\frac{1}{x} = e^{-t}.$$

Step 2: Integration by parts
For $$\int t\,e^{-t}\,dt,$$ choose
$$u = t \;\; \Longrightarrow \;\; du = dt,$$
$$dv = e^{-t}\,dt \;\; \Longrightarrow \;\; v = -e^{-t}.$$

Using the integration-by-parts formula $$\int u\,dv = uv - \int v\,du,$$ we get
$$\int t\,e^{-t}\,dt = (-t\,e^{-t}) - \int (-e^{-t})\,dt = -t\,e^{-t} - \bigl(-e^{-t}\bigr) + C = -t\,e^{-t} - e^{-t} + C.$$

Step 3: Back-substitution
Replace $$t = \ln x$$ and $$e^{-t} = \dfrac{1}{x}:$$
$$I = -(\ln x)\,\frac{1}{x} - \frac{1}{x} + C = -\frac{\ln x}{x} - \frac{1}{x} + C.$$

Hence, the required indefinite integral is
$$\boxed{-\dfrac{\ln x}{x} - \dfrac{1}{x} + C}.$$

Therefore, the correct choice is:
Option A which is: $$-\dfrac{\ln x}{x} - \dfrac{1}{x} + C.$$