Answer
$$ = - \frac{{{{\left( {\coth x} \right)}^3}}}{3} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\coth }^2}x{{\operatorname{csch} }^2}x} dx \cr
& {\text{substitute }}u = \coth x,{\text{ }}du = - {\operatorname{csch} ^2}xdx \cr
& = \int {{{\coth }^2}x{{\operatorname{csch} }^2}x} dx = \int {{u^2}\left( { - du} \right)} \cr
& = - \int {{u^2}du} \cr
& {\text{find the antiderivarive using the power rule}} \cr
& = - \frac{{{u^3}}}{3} + C \cr
& {\text{write in terms of }}x \cr
& = - \frac{{{{\left( {\coth x} \right)}^3}}}{3} + C \cr} $$
