Answer
$\frac{1}{{1 - {e^u}}} + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^u}}}{{{{\left( {1 - {e^u}} \right)}^2}}}} du \cr
& {\text{Let }}z = 1 - {e^u},{\text{ then }}dz = - {e^u}du,{\text{ }} - dz = {e^u}du \cr
& {\text{Apply the substitution}} \cr
& \int {\frac{{{e^u}}}{{{{\left( {1 - {e^u}} \right)}^2}}}} du = \int {\frac{{ - dz}}{{{z^2}}}} \cr
& = - \int {\frac{1}{{{z^2}}}} du \cr
& = - \int {{z^{ - 2}}} du \cr
& {\text{Integrate apply the power rule }}\int {{z^n}dz} = \frac{{{z^{n + 1}}}}{{n + 1}} + C{\text{ }} \cr
& = - \left( {\frac{{{z^{ - 1}}}}{{ - 1}}} \right) + C \cr
& = \frac{1}{z} + C \cr
& {\text{Write in terms of }}u,{\text{ substitute }}z = 1 - {e^u} \cr
& = \frac{1}{{1 - {e^u}}} + C \cr} $$
