Answer
$$\ln \left( {\frac{3}{5}} \right)$$
Work Step by Step
$$\eqalign{
& \int_0^{\ln 3} {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}} dx \cr
& {\text{hyperbolic identity }}\tanh x = \frac{{\sinh x}}{{\cosh x}} = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}} \cr
& = \int_0^{\ln 3} {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}} dx = \int_0^{\ln 3} {\frac{{\sinh x}}{{\cosh x}}} dx \cr
& {\text{find the antiderivative}} \cr
& = \left. {\left( {\ln \cosh x} \right)} \right|_0^{\ln 3} \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = - \left( {\ln \cosh \left( {\ln 3} \right) - \ln \cosh 0} \right) \cr
& {\text{simplifiyng}} \cr
& = - \left( {\ln \left( {\frac{5}{3}} \right) - \ln \left( 1 \right)} \right) \cr
& = - \ln \left( {\frac{5}{3}} \right) \cr
& = \ln \left( {\frac{3}{5}} \right) \cr} $$
