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.

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
$$
\cdots = -\frac{\log(2)}{1} + 2[\arctan(x)]_0^1= -\log(2) + 2( \frac{\pi}{4}-0) = \frac{\pi}{2}-\log(2).
$$
