${\int }_{-\pi }^{\pi } \frac{2x(1+\sin x)}{1+{\cos }^{2}x}dx$  is equal  to

We have, 

$\begin{array}{l}I={\int }_{-\pi }^{\pi } \frac{2x(1+\sin x)}{1+{\cos }^{2}x}dx={\int }_{-\pi }^{\pi }\frac{2x}{1+{\cos }^{2}x}dx+2{\int }_{-\pi }^{\pi } \frac{x\sin x}{1+{\cos }^{2}x}dx\\ \\ \Rightarrow I=0+2\times 2{\int }_0^{\pi } \frac{x\sin x}{1+{\cos }^{2}x}dx\end{array}$

$\begin{array}{l}\Rightarrow I=4{\int }_0^{\pi } \frac{(\pi -x)\sin (\pi -x)}{1+{\cos }^2(\pi -x)}dx\\ \Rightarrow I=4{\int }_0^{\pi } \frac{\pi \sin x}{1+{\cos }^{2}x}dx-4{\int }_0^{\pi } \frac{x\sin x}{1+{\cos }^{2}x}dx\\ \Rightarrow I=-4\pi {\left[{\tan }^{-1}(\cos x)\right]}_0^{\pi }-I\\ \Rightarrow 2I=-4\pi [-\pi /4-\pi /4]\Rightarrow I={\pi }^2\end{array}$