Help to Prove that $\int_{0}^{\pi\over 4}\arctan{(\cot^2{x})}\mathrm dx={2\pi^2-\ln^2({3+2\sqrt{2})}\over 16}$ I need help on proving $(1)$.
$$I=\int_{0}^{\pi\over 4}\arctan{(\cot^2{x})}\mathrm dx={2\pi^2-\ln^2({3+2\sqrt{2})}\over 16}\tag1$$
This is what I have attempted;
Enforcing a sub: $u=\cot^2{x}$ then $du=-2\cot{x}\csc^2{x}dx$
Recall $1+\cot^2{x}=\csc^2{x}$
$$I={1\over2}\int_{1}^{\infty}\arctan{u}\cdot{\mathrm dx\over u^{1/2}+u^{3/2}}$$
Recall $u^3+1=(u+1)(u^2-u+1)$
$$I={1\over2}\int_{1}^{\infty}\arctan{u}\left({A\over u^{1/2}}+{B\over u+1}+{Cu+D\over u^2-u+1}\right)\mathrm du$$
I am stuck at this point.
Can anyone help to prove $(1)$?
 A: We first write $I$ as
$$ I= \frac{\pi^2}{16} + \int_{0}^{\frac{\pi}{4}} \left( \arctan(\cot^2 x) - \arctan(1) \right) \, dx. $$
Now using addition formulas for $\arctan$ and $\cos$, we have
$$ \arctan(\cot^2 x) - \arctan(1)
= \arctan\left(\frac{\cot^2 x - 1}{\cot^2 x + 1} \right)
= \arctan(\cos 2x). $$
Consequently we have
\begin{align*}
I
&= \frac{\pi^2}{16} + \int_{0}^{\frac{\pi}{4}} \arctan(\cos 2x) \, dx \\
&= \frac{\pi^2}{16} + \frac{1}{2}\int_{0}^{\frac{\pi}{2}} \arctan(\sin \theta) \, d\theta,
\end{align*}
where the last line follows from the substitution $\theta = \frac{\pi}{2} - 2x$. The last integral can be computed in terms of the Legendre chi function $\chi_2$:
$$ \int_{0}^{\frac{\pi}{2}} \arctan(\sin \theta) \, d\theta = 2\chi_2(\sqrt{2}-1). \tag{1} $$
For a proof of $\text{(1)}$, see my previous answer for instance. There are only a handful of known special values of $\chi_2$, but thankfully
$$\chi_2(\sqrt{2}-1) = \frac{\pi^2}{16} - \frac{1}{4}\log^2(\sqrt{2}+1) \tag{2} $$
is one of them. Summarizing, we have
$$ I = \frac{\pi^2}{8} - \frac{1}{4}\log^2(\sqrt{2}+1), $$
which coincides with the proposed answer.

Addendum. The identity $\text{(2)}$ follows by plugging $x = \sqrt{2}-1$ to the identity
$$ \chi_2\left(\frac{1-x}{1+x}\right) + \chi_2(x) = \frac{\pi^2}{8} - \frac{1}{2}\log x \log\left(\frac{1-x}{1+x}\right), $$
which can be easily checked by differentiating both sides.
