Answering https://math.stackexchange.com/a/4983355/186817 i have realized that i have proved, $$a\neq 0,\left(\arctan\left(\frac{1}{a}\right)\right)^2=\int_0^1\frac{x\ln\left(\frac{1+x}{1-x}\right)}{a^2+x^2}dx$$
What is your preferred method to compute this?
Proof:
$$ \begin{gathered} \Omega=\arctan \left(\frac{1}{a}\right)^2=2 \int_0^{\frac{1}{a}} \frac{\arctan (y)}{1+y^2} d y=\int_0^1 \int_0^{\frac{1}{a}} \frac{2 y}{\left(1+y^2\right)\left(1+x^2 y^2\right)} d y d x \\ \Omega=\int_0^1 \frac{1}{1-x^2} \int_0^{\frac{1}{a}}\left(\frac{2 y}{1+y^2}-\frac{2 y x^2}{1+x^2 y^2}\right) d y d x=\int_0^1 \frac{1}{1-x^2}\left[\ln \left(\frac{1+y^2}{1+x^2 y^2}\right)\right]_0^{\frac{1}{a}} d x \\ \Omega=\int_0^1 \frac{\ln \left(\frac{1+a^2}{x^2+a^2}\right)}{1-x^2} d x=\left[\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right) \ln \left(\frac{1+a^2}{x^2+a^2}\right)\right]_0^1+\int_0^1 \frac{x \ln \left(\frac{1+x}{1-x}\right)}{a^2+x^2} d x \\ \arctan \left(\frac{1}{a}\right)^2=\int_0^1 \frac{x \ln \left(\frac{1+x}{1-x}\right)}{a^2+x^2} d x \quad \text { Proved } \end{gathered} $$
