Answer
$$\ln \left( {{e^x} + {e^{ - x}}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}} dx \cr
& {\text{set }}u = {e^x} + {e^{ - x}}{\text{ then }}du = \left( {{e^x} - {e^{ - x}}} \right)dx \cr
& {\text{use the change of variable}} \cr
& \int {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}} dx = \int {\frac{{du}}{u}} \cr
& {\text{integrate}} \cr
& = \ln \left| u \right| + C \cr
& {\text{replace }}u = {e^x} + {e^{ - x}} \cr
& = \ln \left| {{e^x} + {e^{ - x}}} \right| + C \cr
& or \cr
& = \ln \left( {{e^x} + {e^{ - x}}} \right) + C \cr} $$
