Answer
$$\ln \left( {1 + \cosh x} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sinh x}}{{1 + \cosh x}}} dx \cr
& {\text{substitute }}u = 1 + \cosh x,{\text{ }}du = \sinh xdx{\text{ }} \cr
& \int {\frac{{\sinh x}}{{1 + \cosh x}}} dx = \int {\frac{{du}}{u}} \cr
& {\text{find the antiderivative}} \cr
& \int {\frac{{du}}{u}} = \ln \left| u \right| + C \cr
& {\text{ with}}\,\,\,u = 1 + \cosh x \cr
& = \ln \left| {1 + \cosh x} \right| + C \cr
& or \cr
& = \ln \left( {1 + \cosh x} \right) + C \cr} $$
