Answer
$$\ln \left( {{e^2} + e + 1} \right)$$
Work Step by Step
$$\eqalign{
& \int_1^3 {\frac{{{e^x}}}{{{e^x} - 1}}} dx \cr
& {\text{Integrate using the substitution method}} \cr
& u = {e^x} - 1,{\text{ }}du = {e^x}dx \cr
& {\text{The new limits of integration are}} \cr
& x = 3 \to u = {e^3} - 1 \cr
& x = 1 \to u = e - 1 \cr
& {\text{Substitute and integrate}} \cr
& \int_1^3 {\frac{{{e^x}}}{{{e^x} - 1}}} dx = \int_{e - 1}^{{e^3} - 1} {\frac{{du}}{u}} \cr
& = \left[ {\ln \left| u \right|} \right]_{e - 1}^{{e^3} - 1} \cr
& = \ln \left| {{e^3} - 1} \right| - \ln \left( {e - 1} \right) \cr
& = \ln \left( {\frac{{{e^3} - 1}}{{e - 1}}} \right) \cr
& = \ln \left( {\frac{{\left( {e - 1} \right)\left( {{e^2} + e + 1} \right)}}{{e - 1}}} \right) \cr
& = \ln \left( {{e^2} + e + 1} \right) \cr} $$
