$\int \frac{x+\sin x}{1+\cos x} dx$ is equal to

$x \tan \frac{x}{2}+c$

$I=\int \frac{x+\sin x}{1+\cos x} d x=\frac{1}{2} \int \sec ^2 \frac{x}{2}(x+\sin x) d x$

$I=\frac{1}{2} \int x \sec ^2 \frac{x}{2}+\frac{2}{2} \int \tan \frac{x}{2} d x$

$=\frac{1}{2} x \frac{\tan x / 2}{1 / 2}-\frac{1}{2} \int \frac{\tan x / 2}{1 / 2}+\int \tan x / 2 d x=x \tan x / 2+c$

Hence (1) is the correct answer.