The teacher hasn't uploaded a solution for this question yet.
We can split the integral into two parts: $$\int\frac{x+\sin x}{1+\cos x}dx = \int\frac{x}{1+\cos x}dx + \int\frac{\sin x}{1+\cos x}dx$$
Using the identities $1+\cos x = 2\cos^2\frac{x}{2}$ and $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$, we get: $$\int\frac{x}{2\cos^2\frac{x}{2}}dx + \int\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}}dx$$ $$\frac{1}{2}\int x\sec^2\frac{x}{2}dx + \int\tan\frac{x}{2}dx$$
For the first integral, we use integration by parts: $\int u dv = uv - \int v du$. Let $u = x$ and $dv = \sec^2\frac{x}{2}dx$. Then $du = dx$ and $v = \int \sec^2\frac{x}{2}dx = 2\tan\frac{x}{2}$. $$\frac{1}{2}\left[x\cdot 2\tan\frac{x}{2} - \int 2\tan\frac{x}{2}dx\right] + \int\tan\frac{x}{2}dx$$ $$x\tan\frac{x}{2} - \int\tan\frac{x}{2}dx + \int\tan\frac{x}{2}dx$$
The two integrals cancel each other out: $$x\tan\frac{x}{2} + C$$
Final Answer: $x\tan\frac{x}{2} + C$
AI generated content. Review strictly for academic accuracy.