Answer
$\int \frac{csc^2~x}{1+cot~x}~dx = -ln \vert 1+cot~x \vert+C$
Work Step by Step
$\int \frac{csc^2~x}{1+cot~x}~dx$
Let $u = cot~x$
$\frac{du}{dx} = -csc^2~x$
$dx = \frac{du}{-csc^2~x}$
$\int \frac{csc^2~x}{1+u} \cdot \frac{du}{-csc^2~x}$
$=\int -\frac{1}{1+u}~du$
$=-ln \vert 1+u \vert+C$
$=-ln \vert 1+cot~x \vert+C$
