The Return II

$\begin{aligned} I & = \int_0^\pi \frac {x\sin x}{1+\cos^2 x} dx & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_0^\pi \left(\frac {x\sin x}{1+\cos^2 x} + \frac {(\pi-x)\sin (\pi-x)}{1+\cos^2 (\pi-x)} \right) dx \\ & = \frac 12 \int_0^\pi \left(\frac {x\sin x}{1+\cos^2 x} + \frac {(\pi-x)\sin x}{1+\cos^2 x} \right) dx \\ & = \frac 12 \int_0^\pi \frac {\pi \sin x}{1+\cos^2 x} dx \\ & = \frac \pi 2 \int_0^\pi \frac {-d \sin x}{1+\cos^2 x} \\ & = \frac \pi 2 \tan^{-1} (\cos x) \bigg|_\pi^0 \\ & = \frac {\pi^2}4 \end{aligned}$

Therefore, $a+b=2+4 = \boxed 6$ .