Answer
$$\operatorname{csch} \left( {\frac{1}{x}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\operatorname{csch} \left( {1/x} \right)\coth \left( {1/x} \right)}}{{{x^2}}}} dx \cr
& {\text{Let }}u = \frac{1}{x},{\text{ }}du = - \frac{1}{{{x^2}}}dx,{\text{ }}dx = - {x^2}du \cr
& {\text{Substituting}} \cr
& \int {\frac{{\operatorname{csch} \left( {1/x} \right)\coth \left( {1/x} \right)}}{{{x^2}}}} dx = \int {\frac{{\operatorname{csch} u\coth u}}{{{x^2}}}} \left( { - {x^2}} \right)du \cr
& = - \int {\operatorname{csch} u\coth u} du \cr
& {\text{Integrating}} \cr
& = - \left( { - \operatorname{csch} u} \right) + C \cr
& = \operatorname{csch} u + C \cr
& {\text{Write in terms of }}x \cr
& = \operatorname{csch} \left( {\frac{1}{x}} \right) + C \cr} $$
