Answer
$$\ln \left( {{e^x} - {e^{ - x}}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^x} + {e^{ - x}}}}{{{e^x} - {e^{ - x}}}}} dx \cr
& {\text{Let }}u = {e^x} - {e^{ - x}},{\text{ }}du = \left( {{e^x} + {e^{ - x}}} \right)dx \cr
& {\text{Apply the substitution}} \cr
& \int {\frac{{{e^x} + {e^{ - x}}}}{{{e^x} - {e^{ - x}}}}} dx = \int {\frac{{du}}{u}} \cr
& {\text{Integrating}} \cr
& {\text{ }} = \ln u + C \cr
& {\text{Back - substitute }}u = {e^x} - {e^{ - x}} \cr
& {\text{ }} = \ln \left( {{e^x} - {e^{ - x}}} \right) + C \cr} $$
