Answer
$$\ln \left| {\frac{{1 + {e^x}}}{{{e^x}}}} \right| - \frac{1}{{{e^x}}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{e^x} + {e^{2x}}}}} \cr
& {\text{set }}u = {e^x},{\text{ }}du = {e^x}dx,{\text{ }}dx = \frac{{du}}{{{e^x}}} = \frac{{du}}{u} \cr
& {\text{substitute}} \cr
& \int {\frac{{dx}}{{{e^x} + {e^{2x}}}}} = \int {\frac{{du/u}}{{u + {u^2}}}} \cr
& = \int {\frac{{du}}{{u\left( {u + {u^2}} \right)}}} \cr
& {\text{partial fractions}} \cr
& \frac{1}{{u\left( {u + {u^2}} \right)}} = \frac{1}{{{u^2}\left( {1 + u} \right)}} = \frac{A}{u} + \frac{B}{{{u^2}}} + \frac{C}{{1 + u}} \cr
& 1 = Au\left( {1 + u} \right) + B\left( {1 + u} \right) + C{u^2} \cr
& 1 = Au + A{u^2} + B + Bu + C{u^2} \cr
& 1 = \left( {A{u^2} + C{u^2}} \right) + Au + Bu + B \cr
& {\text{by equating the coefficients}} \cr
& {u^2}:{\text{ }}A + C = 0 \cr
& u:{\text{ }}A + B = 0 \cr
& {x^0}:{\text{ }}B = 1 \cr
& {\text{solving these equations}} \cr
& A = - 1 \cr
& B = 1 \cr
& C = 1 \cr
& {\text{substituting constants}} \cr
& \frac{1}{{u\left( {u + {u^2}} \right)}} = - \frac{1}{u} + \frac{1}{{{u^2}}} + \frac{1}{{1 + u}} \cr
& = \int {\frac{{du}}{{u\left( {u + {u^2}} \right)}}} = \int {\left( { - \frac{1}{u} + \frac{1}{{{u^2}}} + \frac{1}{{1 + u}}} \right)} du \cr
& = - \ln \left| u \right| + \ln \left| {1 + u} \right| - \frac{1}{u} + C \cr
& = \ln \left| {\frac{{1 + u}}{u}} \right| - \frac{1}{u} + C \cr
& {\text{with }}u = {e^x} \cr
& = \ln \left| {\frac{{1 + {e^x}}}{{{e^x}}}} \right| - \frac{1}{{{e^x}}} + C \cr} $$