$\displaystyle \int^1_0 arcsin\left(\frac{2x}{1+x^2}\right)$
PS. How do you get the brackets to enclose the entire expression?
I was tryin' to find a nice way to tackle this but I couldn't.
Looks like it's pretty straightforward.
Integrate by parts:
$\displaystyle u = \arcsin \frac{{2x}}
{{1 + x^2 }} \implies du = \frac{2}
{{1 + x^2 }}\,dx$ & $\displaystyle dv=dx,$ so
$\displaystyle \int_0^1 {\arcsin \frac{{2x}}
{{1 + x^2 }}\,dx} = \left. {x\arcsin \frac{{2x}}
{{1 + x^2 }}} \right|_0^1 - \int_0^1 {\frac{{2x}}
{{1 + x^2 }}\,dx} .$
The rest follows.