The value of definite integral  $\underset{\text{-π}}{\overset{\text{π}}{\int }}\frac{\text{2x}\left(\text{1+sinx}\right)}{{\text{1+cos}}^{\text{2}}\text{x}}\text{dx}$  is:

$I=\underset{-\pi }{\overset{\pi }{\int }}\frac{2x(1+\sin x)}{1+{\cos }^{2}x}dx$

Using King and add

$$\begin{gathered} 2I=\underset{-\pi }{\overset{\pi }{\int }}\frac{2x(1+\sin x)+2(-x)(1-\sin x)}{1+{\cos }^{2}x}dx=4\underset{-\pi }{\overset{\pi }{\int }}\frac{x\sin x}{1+{\cos }^{2}x}dx \\ I=2\underset{-\pi }{\overset{\pi }{\int }}f\left(x\right)dx \\ I=4\underset{0}{\overset{\pi }{\int }}\frac{x\sin x}{1+{\cos }^{2}x}dx \end{gathered}$$

Using Kind and add

$$\begin{gathered} 2I=4\pi \underset{0}{\overset{\pi }{\int }}\frac{\sin x}{1+{\cos }^{2}x}dx \\ Put \cos x=t \\ 2I=4\pi \underset{-1}{\overset{1}{\int }}\frac{dt}{1+t^2} \\ \therefore I=4\pi \times \frac{\pi }{4}={\pi }^2 \end{gathered}$$