$\int\frac{x}{(x\tan x+1)^2}\,\mathrm{d}x$

Question

$$ \int\frac{x}{(x\tan x+1)^2}\,\mathrm{d}x = \int x \frac{\cot^2 x}{(x+\cot x)^2}\,\mathrm{d}x.$$

By parts method gives $$-\frac{x}{x+\cot x}+\int\frac{1}{x+\cot x}\,\mathrm{d}x,$$ and how to solve it?