Let $z=\pi - x$, therefore $dz = -dx$. It follows that

$$\int_0^{\pi}\frac{x\sin{x}}{1+\cos^2 x}dx = -\int_{\pi}^0\frac{(\pi-z)\sin(\pi - z)}{1+\cos^2(\pi-z)}dz = \int_0^{\pi}\frac{\pi\sin{z}}{1+\cos^2 z}dz - \int_0^{\pi}\frac{z\sin{z}}{1+\cos^2 z}dz$$

Therefore

$$\int_0^{\pi}\frac{x\sin{x}}{1+\cos^2 x}dx=\frac{\pi}{2}\int_0^{\pi}\frac{\sin{x}}{1+\cos^2x}dx=\frac{\pi}{2}\int_0^{\pi}\frac{d(-\cos{x})}{1+\cos^2x}$$

Let $u=-\cos{x}$, then the above integral equals

$$\frac{\pi}{2}\int_{-1}^{1}\frac{du}{u^2}=\frac{\pi}{2}\left.\arctan{u}\right|_{-1}^{1}=\frac{\pi}{2}\left(\frac{\pi}{4}+\frac{\pi}{4}\right)=\boxed{\frac{\pi^2}{4}}$$