Answer
$$\ln \left( {{e^x} + {e^{ - x}}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}dx} \cr
& {\text{substitute }}u = {e^x} + {e^{ - x}}, \cr
& {\text{ }}du = \left( {{e^x} - {e^{ - x}}} \right)dx \cr
& \int {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}dx} = \int {\frac{{du}}{u}} \cr
& {\text{find the antiderivative use the logarithmic rule}} \cr
& = \ln \left| u \right| + C \cr
& {\text{ with}}\,\,\,u = {e^x} + {e^{ - x}} \cr
& = \ln \left| {{e^x} + {e^{ - x}}} \right| + C \cr
& = \ln \left( {{e^x} + {e^{ - x}}} \right) + C \cr} $$
