Use property: $I = \int_{0}^{\pi} f(x)dx = \int_{0}^{\pi} f(\pi - x)dx$ 

 Compute 

$f(\pi - x) = \dfrac{(\pi - x)\sin(\pi - x)}{1 + \cos^2(\pi - x)} = \dfrac{(\pi - x)\sin x}{1 + \cos^2 x}$ 

 Add them: $f(x) + f(\pi - x) = \dfrac{\pi \sin x}{1 + \cos^2 x}$ 

 So, $2I = \displaystyle \int_{0}^{\pi} \frac{\pi \sin x}{1 + \cos^2 x}dx$ 

 Let $u = \cos x$, $du = -\sin x,dx$. 

When $x=0$, $u=1$, and when $x=\pi$, $u=-1$: 

$2I = \pi \displaystyle \int_{1}^{-1} \frac{-du}{1+u^2}$ 

$2I = \pi \displaystyle \int_{-1}^{1} \frac{du}{1+u^2}$ 

 This equals: $2I = \pi\left[\tan^{-1}u\right]_{-1}^{1} = \pi\left(\dfrac{\pi}{4} - \left(-\dfrac{\pi}{4}\right)\right)$ 

$2I = \pi \cdot \dfrac{\pi}{2} = \dfrac{\pi^2}{2}$ 

 So, $I = \dfrac{\pi^2}{4}$