Integral $\int_{0}^{1} \frac{\ln(x^2+1)}{x^2} \,dx$ I would like to ask how to integrate the following integration.
\begin{gather*}
    \int_{0}^{1} \frac{\ln(x^2+1)}{x^2} \,dx
\end{gather*}
I try to use integration by parts and I stuck with the following
\begin{gather*}
   \frac{\ln(x^2+1)}{x} - \int_{0}^{1} d\frac{\ln(x^2+1)}{x^2} \
\end{gather*}
I am confused with the continue step.
 A: Let $u=\log(x^2+1)$ and $dv=\frac{dx}{x^2}$.

Then $du = \frac{2xdx}{x^2+1}$ and $v=\frac{-1}{x}$.
Therefore
$$
\int_0^1 \frac{\log(x^2+1)}{x^2}dx = \left[\frac{-\log(x^2+1)}{x}\right]_0^1 - \int_0^1\frac{2dx}{x^2+1}
$$
The value of $\frac{\log(x^2+1)}{x}$ at 0 is undefined, but we can compute it with de l'Hôpital's rule:
$$
\lim_{x\to0}\frac{\log(x^2+1)}{x} = \lim_{x\to0}\frac{-2x}{x^2+1} = 0.
$$
Plugging everything in, we obtain
$$
\cdots = -\frac{\log(2)}{1} + 2[\arctan(x)]_0^1= -\log(2) + 2( \frac{\pi}{4}-0) = \frac{\pi}{2}-\log(2).
$$
A: As the comments point out, there are a few details you can (and should!) work out yourself, but as a hint, your integration by parts should be (after some simplification)
$$ \lim_{b\to0} \frac{\ln(x^2+1)}{x} \bigg|_{x=1}^{x=b} + 2 \int_0^1 \frac{dx}{1+x^2}. $$
A: Good idea using the integration by part.
Let's recall that integration by Parts is a special method of integration that is often useful when two functions are multiplied together.
At first let us see the rule: $$\int f(x)g(x) dx=u(x)\int v(x) dx-\int u'(x) \Big(\int v(x) dx\Big) dx$$
Now for your problem apply this rule with $v(x)=\frac{1}{x^2}$ and $u(x)=\log{(1+x^2)}$.  So:
$$\int \frac{\log{(1+x^2)}}{x^2}=\log{(1+x^2)}\int \frac{1}{x^2} dx-\int (\log{(1+x^2)})' \Big(\int \frac{1}{x^2} dx\Big) dx=$$ $$=-\frac{1}{x}\log{(1+x^2)}+\int \frac{1}{x}\frac{2x}{1+x^2}=-\frac{1}{x}\ln{(1+x^2)}+2\arctan{(x)}+c$$ Obviously I have shown the indefinite integral but in your case you have to evaluate the integral with the extremes ($0$ and $1$) imposed, in fact your problem I think was the integration itself.
