Answer
$$ x - \ln \left( {1 + {e^x}} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{1 + {e^x}}}} \cr
& {\text{write }}{e^x}{\text{ as }}\frac{1}{{{e^{ - x}}}} \cr
& = \int {\frac{{dx}}{{1 + 1/{e^{ - x}}}}} \cr
& {\text{simplify the integrand}} \cr
& = \int {\frac{{dx}}{{\frac{{{e^{ - x}} + 1}}{{{e^{ - x}}}}}}} \cr
& = \int {\frac{{{e^{ - x}}dx}}{{{e^{ - x}} + 1}}} \cr
& {\text{set }}u = {e^{ - x}} + 1\cr
& {\text{then }}\frac{{du}}{{dx}} = - {e^{ - x}} \to - du = {e^{ - x}}dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{{{e^{ - x}}dx}}{{{e^{ - x}} + 1}}} = \int {\frac{{ - du}}{u}} \cr
& {\text{integrating}} \cr
& = - \ln \left| u \right| + C \cr
& {\text{replace }}{e^{ - x}} + 1{\text{ for }}u \cr
& = - \ln \left| {{e^{ - x}} + 1} \right| + C \cr
& {\text{simplifying, we get:}} \cr
& = - \ln \left( {{e^{ - x}} + 1} \right) + C \cr
& = - \ln \left( {\frac{1}{{{e^x}}} + 1} \right) + C \cr
& = - \ln \left( {\frac{{1 + {e^x}}}{{{e^x}}}} \right) + C \cr
& = \ln \left( {\frac{{{e^x}}}{{1 + {e^x}}}} \right) + C \cr
& = \ln \left( {{e^x}} \right) - \ln \left( {1 + {e^x}} \right) + C \cr
& = x - \ln \left( {1 + {e^x}} \right) + C \cr} $$
