Evaluate the integral $\int_0^{\infty} \left(\frac{\log x \arctan x}{x}\right)^2 \ dx$ Some rumours point out that the integral you see might be evaluated in a 
straightforward way.
But rumours are sometimes just rumours. Could you confirm/refute it?
$$
\int_0^{\infty}\left[\frac{\log\left(x\right)\arctan\left(x\right)}{x}\right]^{2}
\,{\rm d}x
$$
EDIT
W|A tells the integral evaluates $0$ but this is not true.
Then how do I exactly compute it?

Related problems: (I), (II), (III). Denoting our integral by $J$ and recalling the mellin transform
$$ F(s)=\int_{0}^{\infty}x^{s-1} f(x)\,dx  \implies  F''(s)=\int_{0}^{\infty}x^{s-1} \ln(x)^2\,f(x)\,dx.$$
Taking $f(x)=\arctan(x)^2$, then the mellin transform of $f(x)$ is
$$ \frac{1}{2}\,{\frac {\pi \, \left( \gamma+2\,\ln\left( 2 \right) +\psi
\left( \frac{1}{2}+\frac{s}{2} \right)\right) }{s\sin \left( \frac{\pi \,s}{2}
\right)}}-\frac{1}{2}\,{\frac {{\pi }^{2}}{s\cos\left( \frac{\pi \,s}{2}
 \right) }},$$
where $\psi(x)=\frac{d}{dx}\ln \Gamma(x)$ is the digamma function. Thus $J$ can be calculated directly as
$$ J= \lim_{s\to -1} F''(s) = \frac{1}{12}\,\pi \, \left( 3\,{\pi }^{2}\ln  \left( 2 \right) -{\pi }^{2}+24
\,\ln  \left( 2 \right) -3\,\zeta  \left( 3 \right)  \right)\sim 6.200200824 .$$ 
