The value of the integral $\int\limits_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x$, is

$\frac{\pi^2}{4}$

Let $I=\int\limits_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x$     .......(i)

Then,

$\Rightarrow I=\int\limits_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+\cos ^2(\pi-x)} d x$         [Using $\int\limits_0^a f(x)dx = \int\limits_0^a f(a-x) dx$]

$\Rightarrow I=\int\limits_0^\pi \frac{(\pi-x) \sin x}{1+\cos ^2 x} d x$       ......(ii)

Adding (i) and (ii), we get

$2 \pi=\pi \int\limits_0^\pi \frac{\sin x}{1+\cos ^2 x} d x$

$\Rightarrow 2 I=-\pi \int\limits_1^{-1} \frac{1}{1+t^2} d t$, where $t=\cos x$

$\Rightarrow 2 I=-\pi\left[\tan ^{-1} t\right]_1^{-1}=-\pi\left(-\frac{\pi}{4}-\frac{\pi}{4}\right)=\frac{\pi^2}{2} $

$\Rightarrow I=\frac{\pi^2}{4}$