$\int \frac{x+\sin x}{1+\cos x} \mathrm{~d} x=$

The integral $I = \int \frac{x+\sin x}{1+\cos x} dx$ is evaluated using trigonometric identities.

The half-angle identities $1+\cos x = 2\cos^2 \frac{x}{2}$ and $\sin x = 2\sin \frac{x}{2} \cos \frac{x}{2}$ transform the integrand:

$I = \int \frac{x + 2\sin \frac{x}{2} \cos \frac{x}{2}}{2\cos^2 \frac{x}{2}} dx = \int \left( \frac{1}{2} x \sec^2 \frac{x}{2} + \tan \frac{x}{2} \right) dx$

Splitting the integral gives $I = \int \frac{1}{2} x \sec^2 \frac{x}{2} dx + \int \tan \frac{x}{2} dx$.

Integration by parts is applied to the first term, with $u = x$ and $dv = \frac{1}{2} \sec^2 \frac{x}{2} dx$, yielding $v = \tan \frac{x}{2}$:

$\int \frac{1}{2} x \sec^2 \frac{x}{2} dx = x \tan \frac{x}{2} - \int \tan \frac{x}{2} dx$

Substituting back, the terms $\int \tan \frac{x}{2} dx$ cancel, leaving:

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

The result corresponds to option C.