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 ?.

Since you mentioned you tried contour integration, I'll add an approach using contour integration.
Consider $ \displaystyle f(z) = \frac{\log^{2}(1-iz)}{z^{2}}$ where the branch cut runs down the imaginary axis from $z=-i$.
And notice that the singularity at the origin is removable.
Now integrate around a contour that consists of the line segment $[-R,R]$ and the upper half of the circle $|z|=R$.
As $R \to \infty$, $ \displaystyle \int f(z) \ dz$ vanishes along the upper half of $|z|=R$.
Therefore,
$$ \int_{-\infty}^{\infty} \frac{\big( \log|1-ix| - i \arctan (x) \big)^{2}}{x^{2}} \ dx = 0 $$ 
which implies $$\int_{-\infty}^{\infty} \frac{\frac{1}{4} \log^{2}(1+x^{2}) -i \log(1+x^{2}) \arctan (x) - \arctan^{2}(x) }{x^{2}} =0 . $$
And equating the real parts on both sides of the equation we have $$ \begin{align} \int_{-\infty}^{\infty} \frac{\arctan^{2} (x)}{x^{2}} \ dx &= \frac{1}{4} \int_{-\infty}^{\infty} \frac{\log^{2} (1+x^{2})}{x^{2}} \ dx \\ &= \frac{1}{2} \int_{0}^{\infty} \frac{\log^{2}(1+x^{2})}{x^{2}} \ dx \\ &= \frac{1}{4} \int_{0}^{\infty} \frac{\log^{2}(1+u)}{u^{3/2}} \ du \\ &= \frac{1}{4} (8 \pi \log 2)\tag{1} \\ &= 2 \pi \log 2 . \end{align}$$
$(1)$ Integrate $ \int_0^\infty \frac{ \ln^2(1+x)}{x^{3/2}} dx=8\pi \ln 2$
EDIT:
As was stated in Gennaro Marco Devincenzis' answer, integrating by parts shows that $$\int_{-\infty}^{\infty} \frac{\arctan^{2}(x)}{x^{2}} \ dx = 2 \int_{-\infty}^{\infty} \frac{\arctan x}{x(1+x^{2})} \ dx. $$
And that integral as well can be evaluated using contour integration, which is probably preferable since it's just a residue calculation.
See here for the details.
