Answer
$$ - \frac{1}{{12}}{\operatorname{csch} ^4}\left( {3x} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {{{\operatorname{csch} }^4}\left( {3x} \right)\coth \left( {3x} \right)} dx \cr
& {\text{Split, recall that }}{a^m}{a^n} = {a^{m + n}} \cr
& \int {{{\operatorname{csch} }^3}\left( {3x} \right)\operatorname{csch} \left( {3x} \right)\coth \left( {3x} \right)} dx \cr
& {\text{Let }}u = \operatorname{csch} \left( {3x} \right),{\text{ }}du = - \operatorname{csch} \left( {3x} \right)\coth \left( {3x} \right)\left( 3 \right)dx \cr
& - \frac{1}{3}du = \operatorname{csch} \left( {3x} \right)\coth \left( {3x} \right)dx \cr
& {\text{Substitute}} \cr
& \int {{{\operatorname{csch} }^3}\left( {3x} \right)\operatorname{csch} \left( {3x} \right)\coth \left( {3x} \right)} dx = \int {{u^3}\left( { - \frac{1}{3}} \right)} du \cr
& = - \frac{1}{3}\int {{u^3}} du \cr
& {\text{Integrating}} \cr
& = - \frac{1}{3}\left( {\frac{{{u^4}}}{4}} \right) + C \cr
& = - \frac{1}{{12}}{u^4} + C \cr
& {\text{Write in terms of }}x \cr
& = - \frac{1}{{12}}{\operatorname{csch} ^4}\left( {3x} \right) + C \cr} $$
