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

Question

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.

Answer 1

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.

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

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