∫ x + sin x 1 + cos x d x =

Consider

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

$\Rightarrow I=\int \frac{x}{1+\cos x}dx+\int \frac{\sin x}{1+\cos x}dx$

Consider $I=I_1+I_2$

Now,

$I_1=\int \frac{x}{1+\cos x}dx=\int \frac{xdx}{2{\cos }^2{\displaystyle \frac{x}{2}}}$ $\left\{\because 1+\cos \theta =2{\cos }^2\frac{\theta }{2}\right\}$

$\Rightarrow I_1=\int \frac{x}{2}{\sec }^2\frac{x}{2}dx$

$\Rightarrow I_1=2\frac{x}{2}\tan \frac{x}{2}-2{\log }_e\left|\sec \frac{x}{2}\right|+c_1$

$\left\{\because \int uvdx=u\int vdx-\int \left\{\frac{du}{dx}·\int v dx\right\}dx\right\}$

$\Rightarrow I_1=x\tan \frac{x}{2}-{\log }_e\left|{\sec }^2\frac{x}{2}\right|+c_1$

Now,

$I_2=\int \frac{\sin x}{1+\cos x}dx$

Let $1+\cos x=t\Rightarrow -\sin xdx=dt$

$\Rightarrow I_2=-\int \frac{1}{t}dt$

$\Rightarrow I_2=-{\log }_e\left|t\right|+c_2$

$\Rightarrow I_2=-{\log }_e\left|1+\cos x\right|+c_2$

$\Rightarrow I_2=-{\log }_e\left|2{\cos }^2\frac{x}{2}\right|+c_2$

$\Rightarrow I_2=-{\log }_{e}2-{\log }_e\left|{\cos }^2\frac{x}{2}\right|+c_2$

$\Rightarrow I_2={\log }_e\left|{\sec }^2\frac{x}{2}\right|+c_3 ; c_3=c_2-{\log }_{e}2$

So, $I=x\tan \frac{x}{2}-{\log }_e\left|{\sec }^2\frac{x}{2}\right|+c_1+{\log }_e\left|{\sec }^2\frac{x}{2}\right|+c_3$

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