Integrate $\left[\arctan\left(x\right)/x\right]^{2}$ between $-\infty$ and $+\infty$ I have tried to calculate
$$
\int_{-\infty}^{\infty}\left[\arctan\left(x\right) \over x\right]^{2}\,{\rm d}x
$$
with integration by parts and that didn't work. 
I looked up the indefinite integral and found it contained a polylogarithm which I don't know how to use so I tried contour integration  but got stuck.
$${\tt\mbox{Wolfram Alpha said the answer is}}\,\,\,{\large \pi\log\left(4\right)}$$  
Can anyone show me how to do this integral ?.

\begin{align}
\int^{\infty}_{-\infty}\frac{(\arctan{x})^2}{x^2}dx 
&=2\int^{\infty}_{0}\frac{(\arctan{x})^2}{x^2}dx\tag1\\
&=2\int^{\pi/2}_0u^2\csc^2{u} \ du\tag2\\
&=4\int^{\pi/2}_0u\cot{u} \ du\tag3\\
&=-4\int^{\pi/2}_0\ln\sin{u} \ du\tag4\\
&=2\pi\ln{2}\tag5\\
\end{align}
Explanation: 
$1)$ Integrand is even 
$2)$ Use the substitution $x=\tan{u}$ 
$3), 4)$ Integrate by parts 
$5)$ See here 
