Answer
$$\frac{1}{2}\coth \left( {\frac{2}{x}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\operatorname{csch} }^2}\left( {2/x} \right)}}{{{x^2}}}} dx \cr
& {\text{substitute }}u = \frac{2}{x},{\text{ }}du = - \frac{2}{{{x^2}}}dx,\, - \frac{1}{2}du = \frac{{dx}}{{{x^2}}} \cr
& = \int {{{\operatorname{csch} }^2}\left( {2/x} \right)\frac{1}{{{x^2}}}} dx \cr
& = \int {{{\operatorname{csch} }^2}\left( u \right)} \left( { - \frac{1}{2}du} \right) \cr
& = - \frac{1}{2}\int {{{\operatorname{csch} }^2}\left( u \right)} du \cr
& {\text{find the antiderivative }}\left( {{\text{formula 18 page 489}}} \right) \cr
& = - \frac{1}{2}\left( { - \coth u} \right) + C \cr
& = \frac{1}{2}\coth u + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \frac{2}{x} \cr
& = \frac{1}{2}\coth \left( {\frac{2}{x}} \right) + C \cr} $$
