Answer
$$ - \operatorname{sech} x + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sinh x}}{{1 + {{\sinh }^2}x}}dx} \cr
& {\text{From the table of hyperbolic identities page 384}} \cr
& {\text{cos}}{{\text{h}}^2}x - {\sinh ^2}x = 1{\text{ }} \to {\text{ }}{\sinh ^2}x = {\text{cos}}{{\text{h}}^2}x - 1 \cr
& \int {\frac{{\sinh x}}{{1 + {{\sinh }^2}x}}dx} = \int {\frac{{\sinh x}}{{1 + {\text{cos}}{{\text{h}}^2}x - 1}}dx} \cr
& = \int {\frac{{\sinh x}}{{{\text{cos}}{{\text{h}}^2}x}}dx} \cr
& = \int {\left( {\frac{1}{{\cosh x}}} \right)\left( {\frac{{\sinh x}}{{{\text{cosh}}x}}} \right)dx} \cr
& = \int {\operatorname{sech} x\tanh xdx} \cr
& {\text{Integrating }} \cr
& = - \operatorname{sech} x + C \cr} $$
