integral_pi^0(x*sinx)/(1+cosx^2) dx =
integral_6^4(x)/(x-1)*(x+2)*(x-3)dx =
$\displaystyle I = \int_0^\pi \frac{x\sin(x)}{1+\cos^2(x)} ~dx$
Sub $\displaystyle x = \pi - t $
$\displaystyle I = \pi \int_0^{\pi} \frac{\sin(x)}{1+\cos^2(x)}~dx - I $
$\displaystyle 2I = \pi \cdot [ -\arctan(\cos(x)) ]_0^{\pi} = \pi^2/2$
$\displaystyle I = \frac{\pi^2}{4}$