Evaluating $\int_0^{+\infty}(\frac{\arctan t}{t})^2dt$ Is it possible to calculate $$\int_0^{+\infty}\Big(\frac{\arctan t}{t}\Big)^2 dt$$ without using complex analysis? 
I found this on a calculus I book and I don't know how to solve it. 
I tried to set $t = \tan u$ but it didn't help. 

$J=\displaystyle\int_0^{\infty}  \dfrac{dt}{(1+a^2 t^2)(1+b^2 t^2)}=\int_0^{\infty}\dfrac{b^2}{(b^2-a^2)(1+b^2t^2)}dt-\int_0^{\infty}\dfrac{a^2}{(b^2-a^2)(1+a^2t^2)}dt$
If $a>0$, one obtains (change of variable $y=at$):
$\displaystyle \int_0^{\infty}\dfrac{a}{1+a^2t^2}dx=\dfrac{\pi}{2}$
Thus, $\displaystyle J=\dfrac{\pi}{2}\left(\dfrac{b}{b^2-a^2}-\dfrac{a}{b^2-a^2}\right)=\dfrac{\pi}{2(a+b)}$
