Answer
$$\ln \left( {1 + {e^x}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^x}}}{{1 + {e^x}}}} dx;\,\,\,u = 1 + {e^x} \cr
& u = 1 + {e^x} \to du = {e^x}dx,\,\,dx = \frac{1}{{{e^x}}}du \cr
& {\text{using the indicated substitution in the book }} \cr
& \int {\frac{{{e^x}}}{{1 + {e^x}}}} dx = \int {\frac{{{e^x}}}{u}} \left( {\frac{1}{{{e^x}}}du} \right) \cr
& = \int {\frac{1}{u}} du \cr
& {\text{integrate}} \cr
& = \ln \left| u \right| + C \cr
& {\text{substituting }}u = 1 + {e^x} \cr
& = \ln \left| {1 + {e^x}} \right| + C \cr
& 1 + {e^x}{\text{ is always positive}}{\text{, thus}} \cr
& = \ln \left( {1 + {e^x}} \right) + C \cr} $$
