$\int \left(\log \right(\sin x)+x\cot x)dx=$

The given integral is $I=\int \left(\log \right(\sin x)+x\cot x)dx$

$\Rightarrow I=\int \log \left(\sin x\right)dx+\int x\cot xdx \dots \left(1\right)$

Let, $I_1=\int \log \left(\sin x\right)·1dx$, integrating w.r.t. $x$, treating $1$ as second function, we get, 

$I_1=x·\log \left(\sin x\right)-\int \frac{\cos x}{\sin x}·xdx+c=x·\log \left(\sin x\right)-\int \cot x·xdx+c$, where $c$ is the constant of integration.

$\Rightarrow I_1+I_2=I=x\log \left(\sin x\right)+c$