Answer
$$\ln \left( {1 + {e^r}} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^r}}}{{1 + {e^r}}}} dr \cr
& {\text{set }}u = 1 + {e^r}{\text{ then }}\frac{{du}}{{dr}} = {e^r} \to du = {e^r}dr \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{{{e^r}}}{{1 + {e^r}}}} dr = \int {\frac{{du}}{u}} \cr
& {\text{integrating}} \cr
& = \ln \left| u \right| + C \cr
& {\text{replace }}1 + {e^r}{\text{ for }}u \cr
& = \ln \left| {1 + {e^r}} \right| + C \cr
& {\text{or}} \cr
& = \ln \left( {1 + {e^r}} \right) + C \cr} $$
