Answer
$$\ln \left| {1 - {e^{ - s}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{ds}}{{{e^s} - 1}}} \cr
& {\text{Multiply the numerator and denominator by }}{e^{ - s}} \cr
& = \int {\frac{{{e^{ - s}}ds}}{{{e^{ - s}}\left( {{e^s} - 1} \right)}}} \cr
& = \int {\frac{{{e^{ - s}}ds}}{{1 - {e^{ - s}}}}} \cr
& {\text{Integrate by using the substitution method}} \cr
& {\text{Let }}u = 1 - {e^{ - s}},\,\,\,\,du = {e^{ - s}}ds \cr
& {\text{Write the integral in terms of }}u \cr
& \int {\frac{{{e^{ - s}}ds}}{{1 - {e^{ - s}}}}} = \int {\frac{{du}}{u}} \cr
& {\text{Integrate}} \cr
& = \ln \left| u \right| + C \cr
& \cr
& {\text{Write in terms of }}s,{\text{ replace }}1 - {e^{ - s}}{\text{ for }}u \cr
& = \ln \left| {1 - {e^{ - s}}} \right| + C \cr} $$
