Another math contest problem: $\int_0^{\frac{\ln^22}4}\,\frac{\arccos\frac{\exp\sqrt x}{\sqrt2}}{1-\exp\sqrt{4\,x}}dx$ 
Prove:
  $$
{\Large\int_{0}^{\ln^{2}\left(2\right) \over4}}\,
\frac{\arccos\left(\vphantom{\huge A}
{\exp\left(\vphantom{\large A}\sqrt{x\,}\right)
\over
\sqrt{\vphantom{\large A}2\,}}\right)}
{1-\exp\left(\sqrt{4x\,}\,\right)}
\,{\rm d}x
=
-\,\frac{\,\,\pi^{3}}{192}
$$

I haven't solved it yet.
 A: 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:
$$ \exp\sqrt{4x} = t, \qquad t = 2\cos^{2}u, \qquad x = \tan u $$
A: 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 !
