$\int \dfrac {x+\sin }{1+\cos x}dx=$

$\displaystyle \int \dfrac {x+\sin x}{(1+\cos x)}dx=\displaystyle \int \dfrac {x+2 \sin (x/2)\cos (x/2)}{1+2\cos^2 (x/2)-1}dx$

$\displaystyle \int \dfrac {x+2\sin \left (\dfrac {x}{2}\right) \cos \dfrac {x}{2}}{2\cos^2x/2}dx =\displaystyle \int \dfrac {xdx}{2\cos^2 (x/2)}+\displaystyle \int tan \dfrac {x}{2}dx$

$\dfrac { 1 }{ 2 } \int { x } \sec ^{ 2 }{ \left( \dfrac { x }{ 2 } \right) dx } +\int { \tan { \left( \dfrac { x }{ 2 } \right) } dx }$

integrate first integral by parts

$U=x$

$du=dx$

$du=\dfrac {1}{2}\sec^2 x/2 dx$

$u=\tan x/2$

$\dfrac {1}{2} \displaystyle \int x\sec^2 \left (\dfrac {x}{2}\right)dx +\displaystyle \int \tan \dfrac {x}{2}dx =x \tan \dfrac {x}{2}-\displaystyle \int \tan \dfrac {x}{2}dx +\displaystyle \int \tan \dfrac {x}{2}dx$

$=x\tan (x/2)+c$