Evaluating $\int_0^{\sqrt 3} \frac{1}{1+x^2} \sin^{-1} (\frac{2x}{1+x^2}) dx$ The question is to evaluate $$\int_0^{\sqrt 3}  \frac{1}{1+x^2} \sin^{-1} (\frac{2x}{1+x^2}) dx$$
I used the substitution $x=\tan a$ and it led me to the answer $\pi^2 /9$.However the correct answer is $\frac{7 \pi^2}{72}$.I couldn't get why the substitution is leading to a wrong answer.Any ideas?

$x=\tan y\implies0\le y\le\dfrac\pi3$
Using Principal values
$$\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)=\sin^{-1}(\sin2y)=\begin{cases}2y&\mbox{if }0\le2y\le\dfrac\pi2 \\ 
\pi-2y & \mbox{if } \dfrac\pi2 \le2y\le\pi \end{cases}$$
