$\begin{aligned} & \text { Let } I=\int \frac{x+\sin x}{1+\cos x} d x \\ & =\int \frac{x}{1+\cos x} d x+\int \frac{\sin x}{1+\cos x} d x \\ & =\int \frac{x}{2 \cos ^2 \frac{x}{2}} d x+\int \frac{\sin x}{2 \cos ^2 \frac{x}{2}} d x \\ & =\frac{1}{2} \int x \sec ^2 \frac{x}{2} d x+\int \frac{2 \sin x / 2 \cos x / 2}{2 \cos ^2 x / 2} d x \\ & =\frac{1}{2} \int x \sec ^2 \frac{x}{2} d x+\int \tan \frac{x}{2} d x \\ & =\frac{1}{2}\left[x \frac{\tan \frac{x}{2}}{1 / 2}-\int \frac{\tan x / 2}{\frac{1}{2}} d x\right]+\int \tan \frac{x}{2} d x \\ & =x \tan \frac{x}{2}-\int \tan \frac{x}{2} d x+\int \tan \frac{x}{2} d x=x \tan \frac{x}{2}+C\end{aligned}$