Answer
$$ = - \frac{1}{3}\coth 3x + C$$
Work Step by Step
$$\eqalign{
& \int {{{\operatorname{csch} }^2}\left( {3x} \right)} dx \cr
& {\text{substitute }}u = 3x,{\text{ }}du = 3dx \cr
& = \int {{{\operatorname{csch} }^2}\left( {3x} \right)} dx = \int {{{\operatorname{csch} }^2}u\left( {\frac{1}{3}du} \right)} \cr
& = \frac{1}{3}\int {{{\operatorname{csch} }^2}udu} \cr
& {\text{find the antiderivarive using the theorem 6}}{\text{.8}}{\text{.3 }}\left( {{\text{see page 476}}} \right) \cr
& = \frac{1}{3}\left( { - \coth u} \right) + C \cr
& = - \frac{1}{3}\coth u + C \cr
& {\text{write in terms of }}x \cr
& = - \frac{1}{3}\coth 3x + C \cr} $$
