<s>[INST] How to calculate this hard integral $\int_0^{\infty} \frac{\arctan(x)\sqrt{\sqrt{1+x^2}-1}}{x\sqrt{1+x^2}}\,dx$? How to prove that $\displaystyle \int_0^{\infty} \dfrac{\arctan(x)\sqrt{\sqrt{1+x^2}-1}}{x\sqrt{1+x^2}}\,dx= \frac{1}{4}\pi^2\sqrt{2}-\sqrt{2}\ln^2\left(1+\sqrt{2}\right)$ ?
It's very difficult and I have no idea.
 [/INST] $\begin{align}I&=\int_0^{\infty} \dfrac{\arctan(x)\sqrt{\sqrt{1+x^2}-1}}{x\sqrt{1+x^2}}\,dx
\end{align}$
Perform the change of variable $x=\dfrac{2y}{1-y^2}$,
$\begin{align}I&=\sqrt{2}\int_0^{1} \dfrac{\arctan\left(\dfrac{2y}{1-y^2}\right)}{\sqrt{1-y^2}}\,dy\\
&=2\sqrt{2}\int_0^{1} \dfrac{\arctan y}{\sqrt{1-y^2}}\,dy\\
\end{align}$
If one assume that,
$\displaystyle \int_0^{1} \dfrac{\arctan y}{\sqrt{1-y^2}}\,dy=\frac{1}{8}\pi^2-\frac{1}{2}\ln^2(\sqrt{2}+1)$
therefore,
$\displaystyle \boxed{I=\frac{1}{4}\pi^2\sqrt{2}-\sqrt{2}\ln^2(\sqrt{2}+1)}$
(for a proof of this, see: Evaluating a sum involving binomial coefficient in denominator )
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