<s>[INST] Help on this integral $I=\int_0^1 \frac{x \arctan(x)}{1-x^2}\ln\left(\frac{2}{1+x^2}\right) dx$ $$I=\int_0^1 \frac{x \arctan(x)}{1-x^2}\ln\left(\frac{2}{1+x^2}\right) dx$$
Here is my attempt
$\frac{x}{1-x^2}dx=-\frac{1}{2}d\ln(1-x^2)$, integration by part, we got
$$I=-\frac{1}{2}P-Q$$
Where $P=\int_0^1 \frac{\ln(1-x^2)\ln\left(\frac{1+x^2}{2}\right)}{1+x^2}dx$, and
$Q=\int_0^1 \frac{x\cdot\arctan(x)\cdot\ln(1-x^2)}{1+x^2}dx$
How to proceed then? or should I switch to other ways?
 [/INST] Here is a more general solution. Use the contour setup by Sangchul Lee, it can be shown that
$$\int_0^1 \frac{x \arctan a x}{1-x^2}\ln\frac{1+a^2}{1+a^2x^2}dx=\frac13 \arctan^3 a
$$
Let $a=1$ to obtain
$$\int_0^1 \frac{x \arctan x}{1-x^2}\ln\frac{2}{1+x^2}dx=\frac13 \arctan^3 (1)=\frac{\pi^3}{192}
$$
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