Answer
$$\frac{1}{5}\ln \left| {\frac{{{e^x} - 1}}{{{e^x} + 4}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^x}}}{{\left( {{e^x} - 1} \right)\left( {{e^x} + 4} \right)}}} dx \cr
& {\text{Let }}t = {e^x},{\text{ }}dt = {e^x}dx \cr
& {\text{Substituting}} \cr
& \int {\frac{{{e^x}}}{{\left( {{e^x} - 1} \right)\left( {{e^x} + 4} \right)}}} dx = \int {\frac{1}{{\left( {t - 1} \right)\left( {t + 4} \right)}}} dt \cr
& {\text{Decompose into partial fractions}} \cr
& \frac{1}{{\left( {t - 1} \right)\left( {t + 4} \right)}} = \frac{A}{{t - 1}} + \frac{B}{{t + 4}} \cr
& 1 = A\left( {t + 4} \right) + B\left( {t - 1} \right) \cr
& t = 1 \to A = 1/5 \cr
& t = - 4 \to B = - 1/5 \cr
& \frac{1}{{\left( {t - 1} \right)\left( {t + 4} \right)}} = \frac{{1/5}}{{t - 1}} + \frac{{ - 1/5}}{{t + 4}} \cr
& \int {\frac{1}{{\left( {t - 1} \right)\left( {t + 4} \right)}}} dt = \int {\left( {\frac{{1/5}}{{t - 1}} + \frac{{ - 1/5}}{{t + 4}}} \right)} dt \cr
& {\text{Integrate}} \cr
& {\text{ = }}\frac{1}{5}\ln \left| {t - 1} \right| - \frac{1}{5}\ln \left| {t + 4} \right| \cr
& {\text{By using logarithmic properties}} \cr
& = \frac{1}{5}\ln \left| {\frac{{t - 1}}{{t + 4}}} \right| + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{1}{5}\ln \left| {\frac{{{e^x} - 1}}{{{e^x} + 4}}} \right| + C \cr} $$
