$\int \frac{x+\sin x}{1+\cos x}dx=$

$\int \frac{x+\sin x}{1+\cos x}dx$
$\begin{array}{l}=\int \frac{x}{1+\cos x}+\frac{\sin x}{1+\cos x}dx\\ =\int \frac{xdx}{2{\cos }^2\frac{x}{2}}+\int \frac{2\sin x/2\cos {\displaystyle \frac{x}{2}}}{2{\cos }^2\frac{x}{2}}dx\end{array}$
$=\frac{1}{2}\int x{\sec }^2\frac{x}{2}dx+\int \mathrm{Tan}\frac{x}{2}dx$
$=\frac{1}{2}\left[x\int {\sec }^2\frac{x}{2}dx\left.-\int \frac{\tan {\displaystyle \frac{x}{2}}}{1/2}dx\right]\right]+\int \tan \frac{x}{2}dx$
$=\frac{1}{2}x\frac{\tan \frac{x}{2}}{1/2}+c-\int \tan \frac{x}{2}dx+\int \tan \frac{x}{2}dx.$
$=x\tan \frac{x}{2}+C\text{. }$