Answer
$$\frac{1}{3}\ln \left| {\frac{{{e^x} - 1}}{{{e^x} + 2}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^x}}}{{\left( {{e^x} - 1} \right)\left( {{e^x} + 2} \right)}}dx} \cr
& {\text{set }}u = {e^x},{\text{ }}du = {e^x}dx \cr
& \int {\frac{{{e^x}}}{{\left( {{e^x} - 1} \right)\left( {{e^x} + 2} \right)}}dx} = \int {\frac{{du}}{{\left( {u - 1} \right)\left( {u + 2} \right)}}} \cr
& {\text{partial fractions}} \cr
& \frac{1}{{\left( {u - 1} \right)\left( {u + 2} \right)}} = \frac{A}{{u - 1}} + \frac{B}{{u + 2}} \cr
& 1 = A\left( {u + 2} \right) + B\left( {u - 1} \right) \cr
& u = 1 \to \frac{1}{3} = A \cr
& u = - 2 \to - \frac{1}{3} = B \cr
& \frac{A}{{u - 1}} + \frac{B}{{u + 2}} = \frac{{1/3}}{{u - 1}} + \frac{{ - 1/3}}{{u + 2}} \cr
& = \int {\frac{{du}}{{\left( {u - 1} \right)\left( {u + 2} \right)}}} = \int {\left( {\frac{{1/3}}{{u - 1}} + \frac{{ - 1/3}}{{u + 2}}} \right)du} \cr
& {\text{find the antiderivative}} \cr
& = \frac{1}{3}\ln \left| {u - 1} \right| - \frac{1}{3}\ln \left| {u + 2} \right| + C \cr
& = \frac{1}{3}\ln \left| {\frac{{u - 1}}{{u + 2}}} \right| + C \cr
& = \frac{1}{3}\ln \left| {\frac{{{e^x} - 1}}{{{e^x} + 2}}} \right| + C \cr} $$
