Answer
$$ - \frac{{{{\cot }^3}x}}{3} + \cot x + x + C$$
Work Step by Step
$$\eqalign{
& \int {{{\cot }^4}x} dx \cr
& {\text{write as}} \cr
& = \int {{{\cot }^2}x{{\cot }^2}x} dx \cr
& {\text{Use the pythagorean identity }}{\cot ^2}x = {\csc ^2}x - 1 \cr
& = \int {{{\cot }^2}x\left( {{{\csc }^2}x - 1} \right)} dx \cr
& = \int {\left( {{{\cot }^2}x{{\csc }^2}x - {{\cot }^2}x} \right)} dx \cr
& = \int {{{\cot }^2}x{{\csc }^2}x} dx - \int {{{\cot }^2}x} dx \cr
& = \int {{{\cot }^2}x{{\csc }^2}x} dx - \int {\left( {{{\csc }^2}x - 1} \right)} dx \cr
& = \int {{{\cot }^2}x{{\csc }^2}x} dx - \int {{{\csc }^2}x} dx + \int {dx} \cr
& {\text{Integrate}} \cr
& = - \frac{1}{3}{\cot ^3}x - \left( { - \cot x} \right) + x + C \cr
& = - \frac{{{{\cot }^3}x}}{3} + \cot x + x + C \cr} $$
