Answer
$$x - \ln \left| {1 + {e^x}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{1 + {e^x}}}} \cr
& {\text{set }}u = {e^x},{\text{ }}du = {e^x}dx,{\text{ }}dx = \frac{{du}}{{{e^x}}} = \frac{{du}}{u} \cr
& {\text{substitute}} \cr
& \int {\frac{{dx}}{{1 + {e^x}}}} = \int {\frac{{du/u}}{{1 + u}}} \cr
& = \int {\frac{{du}}{{u\left( {1 + u} \right)}}} \cr
& {\text{partial fractions}} \cr
& \frac{1}{{u\left( {1 + u} \right)}} = \frac{A}{u} + \frac{B}{{1 + u}} \cr
& 1 = A\left( {1 + u} \right) + Bu \cr
& u = 0 \to A = 1 \cr
& u = - 1 \to B = - 1 \cr
& \frac{A}{u} + \frac{B}{{1 + u}} = \frac{1}{u} - \frac{1}{{1 + u}} \cr
& = \int {\frac{{du}}{{u\left( {1 + u} \right)}}} = \int {\left( {\frac{1}{u} - \frac{1}{{1 + u}}} \right)} du \cr
& = \ln \left| u \right| - \ln \left| {1 + u} \right| + C \cr
& {\text{with }}u = {e^x} \cr
& = \ln \left| {{e^x}} \right| - \ln \left| {1 + {e^x}} \right| + C \cr
& = x - \ln \left| {1 + {e^x}} \right| + C \cr} $$
