Evaluate the following integral: ∫ ((cot x + cot³ x)/(1 + cot³ x)) dx.

$\int\frac{\cot x + \cot^3 x}{1 + \cot^3 x}dx = \int\frac{\cot x\,\operatorname{cosec}^2 x}{1 + \cot^3 x}dx$ Let $\cot x = t$, so $\operatorname{cosec}^2 x\,dx = -dt$. $= -\int\frac{t\,dt}{1+t^3} = -\int\frac{t\,dt}{(1+t)(t^2-t+1)}$ $\frac{t}{(1+t)(t^2-t+1)} = -\frac{1}{3(t+1)} + \frac{t+1}{3(t^2-t+1)}$, so $\int\frac{t\,dt}{1+t^3} = -\frac13\log(t+1) + \frac16\log(t^2-t+1) + \frac{1}{\sqrt3}\tan^{-1}\frac{2t-1}{\sqrt3}$ Therefore the integral is $\frac13\log(1+\cot x) - \frac16\log(\cot^2 x - \cot x + 1) - \frac{1}{\sqrt3}\tan^{-1}\left(\frac{2\cot x - 1}{\sqrt3}\right) + C$.