Another math contest problem: $\int_0^{\frac{\ln^22}4}\,\frac{\arccos\frac{\exp\sqrt x}{\sqrt2}}{1-\exp\sqrt{4\,x}}dx$

With the substitution $x \mapsto \frac{1}{4}\log^{2}\left(\frac{2}{x^{2}+1}\right)$, it follows that

\begin{align*} \int_{0}^{\frac{\log^{2}2}{4}} \frac{\arccos\left( \frac{\exp\sqrt{x}}{\sqrt{2}} \right)}{1 - \exp\sqrt{4x}} \, dx &= \int_{0}^{1} \frac{x \arctan x}{1 - x^{2}} \log\left(\frac{1+x^{2}}{2}\right) \, dx \\ &= -\frac{1}{2} \int_{-1}^{1} \frac{\arctan x}{1 + x} \log\left(\frac{1+x^{2}}{2}\right) \, dx. \end{align*}

Now you can refer to this solution.

Actually, I obtained this integral representation by applying the following chain of much human-friendly substitutions:

$$ \exp\sqrt{4x} = t, \qquad t = 2\cos^{2}u, \qquad x = \tan u $$

use the residue theorem !

we first assume that :

$\exp\sqrt{x\,}=z$

then your formula turn out to be :

$ {\Large\int_{1}^{\sqrt{2}}}\, \frac{\arccos\left(\vphantom{\huge A} {z \over \sqrt{\vphantom{\large A}2\,}}\right)} {1-z^2} \,{\rm d}x = -\,\frac{\,\,\pi^{3}}{192} $

the residue theorem gives ,

$ {\Large\int_{1}^{\sqrt{2}}}\, \frac{\arccos\left(\vphantom{\ huge A} {z \over \sqrt{\vphantom{\large A}2\,}}\right)} {1-z^2} \,{\rm d}x =\frac{b\pi}{2}\cdot\frac{\arccos(1/\sqrt{2})}{2}\cdot\frac{\arccos(-1/\sqrt{2})}{-2}$$=-\frac{b\pi}{2}\cdot(\frac{\pi}{8})^2$

the last step is to calculate the radius :

$\frac{1}{b}=\frac{z}{2(Inz)^{'}}=$$z^{2}=z_{1}^{2}+z_{2}^{2}=\frac{3}{2}$$\Longrightarrow$$b=\frac{2}{3}$

then, your solution holds !