Answer
$$ - \operatorname{csch} z + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\cosh z}}{{{{\sinh }^2}z}}} dz \cr
& {\text{we can write the integrand as }} \cr
& = \int {\left( {\frac{1}{{\sinh z}}} \right)\left( {\frac{{\cosh z}}{{\sinh z}}} \right)} dz \cr
& {\text{using the hyperbolic identity }}\operatorname{csch} x = \frac{1}{{\sinh x}}{\text{ and }}\coth x = \frac{{\cosh x}}{{\sinh x}} \cr
& = \int {\operatorname{csch} z\coth z} dz \cr
& {\text{integrate using the theorem 6}}{\text{.8 }}\left( {{\text{formula 6}}} \right) \cr
& = \int {\operatorname{csch} z\coth z} dz = - \operatorname{csch} z + C \cr} $$
