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 ?.

You can use the following way to evaluate. It is pretty neat and simple. Let
$$ I(a,b)=\int_{-\infty}^\infty\frac{\arctan(ax)\arctan(bx)}{x^2}dx. $$
Clearly $I(0,b)=I(a,0)=0$ and $I(1,1)=I$. Now
\begin{eqnarray}
\frac{\partial^2I(a,b)}{\partial a\partial b}&=&\int_{-\infty}^\infty\frac{1}{(1+a^2x^2)(1+b^2x^2)}dx\\
&=&\frac{1}{a^2-b^2}\int_{-\infty}^\infty\left(\frac{a^2}{1+a^2x^2}-\frac{b^2}{1+b^2x^2}\right)dx\\
&=&\frac{1}{b^2-a^2}\pi(a-b)\\
&=&\frac{\pi}{a+b}.
\end{eqnarray}
Hence
$$ I=\int_0^1\int_0^1\frac{\pi}{a+b}dadb=2\pi\ln2.$$
