Evaluate the following integrals:

$=\int \frac{\cot x\left(1+\cot ^{2} x\right)}{1+\cot ^{3} x} d x=\int \frac{\cot x \operatorname{cosec}^{2} x}{1+\cot ^{3} x} d x$

Put $\cot x=t,-\operatorname{cosec}^{2} x d x=d t$

$=-\int \frac{t d t}{t^{3}+1}=-\int \frac{t d t}{(t+1)\left(t^{2}-t+1\right)}$

By partial fractions it's a remembering thing

That if you see the above integral just apply the below return result,

$=-\int\left[\frac{(t+1)}{3\left(t^{2}-t+1\right)}-\frac{1}{3(t+1)}\right] d t$

$=\frac{1}{3} \log (t+1)-\frac{1}{3} \int\left[\frac{2 t-1}{2\left(t^{2}-t+1\right)}+\frac{3}{2\left(t^{2}-t+1\right)}\right] d t$

$=\frac{1}{3} \log (t+1)-\frac{1}{6} \log \left(t^{2}-t+1\right)-\frac{1}{2} \int \frac{d t}{\left(t-\frac{1}{2}\right)^{2}+\frac{3}{4}}$

$=\frac{1}{3} \log (t+1)-\frac{1}{6} \log \left(t^{2}-t+1\right)-\frac{1}{2}\left[\frac{2}{\sqrt{3}} \tan ^{-1} \frac{(2 t-1)}{\sqrt{3}}\right]+c$

$=\frac{1}{3} \log (\cot x+1)-\frac{1}{6} \log \left(\cot ^{2} x-\cot x+1\right)-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \cot x-1}{\sqrt{3}}\right)+c$