integration - Evaluation of the integral $int{frac{x+sin{x}}{1+cos{x}}mathrm{d}x}$ by parts

I have to evaluate the following integral by parts: $$\int {\dfrac{x+\sin{x}}{1+\cos{x}}}\mathrm{d}x $$

So I tried to put:

$ u = x + \sin{x}$ $~\qquad\rightarrow \quad$ $\mathrm{d}u=\left(1+\cos{x}\right) \mathrm{d}x$

$\mathrm{d}v = \dfrac{\mathrm{d}x}{1+\cos(x)}$ $\quad \rightarrow \quad$ $v = \int{\dfrac{\mathrm{d}x}{1+\cos{x}}}$

But there is an extra integral to do ( the $v$ function) I evaluated it by sibstitution, and I get $v = \tan{\dfrac{x}{2}}$, Now

$$\int {\dfrac{x+\sin(x)}{1+\cos(x)}}\mathrm{d}x = (x+\sin{x})\, \tan{\dfrac{x}{2}}-x +C$$
My question: is it possible to evaluate this integral entirely by parts (without using any substitution) ?

I appreciate any ideas