<s>[INST] Find integral $\int_0^1 \frac{\ln(1+x^2)}{1+x^2} \ dx$ (most likely substitution) $$\int_0^1 \frac{\ln(1+x^2)}{1+x^2} \ dx$$
I tried letting $x^2=\tan \theta$ but it didn't work. What should I do?
Please don't give full solution, just a hint and I will continue.
 [/INST] Integrate as follows
\begin{align}
&\int_0^1\frac{\ln(1+x^2)}{1+x^2} dx\\ 
=&\int_0^1\int_0^1 \frac{2x^2y}{(1+x^2)(1+x^2y^2)}dy\ dx=\int_0^1\frac{\frac\pi2 y-2\tan^{-1}y}{y^2-1}
\ \overset{y\to \frac{1-y}{1+y}}{dy}\\
= &\ \frac\pi2\int_0^1 \frac{1}{1+y}dy
 -\int_0^1\frac{\tan^{-1}y}{y}dy=\frac\pi2\ln2-G
\end{align}
</s>