Answer
$$\ln \left( {\frac{5}{4}} \right)$$
Work Step by Step
$$\eqalign{
& \int_0^{\ln 2} {\tanh x} dx \cr
& {\text{Use the hyperbolic identity }}\tanh x = \frac{{\sinh x}}{{\cosh x}} \cr
& \int {\tanh x} dx = \int {\frac{{\sinh x}}{{\cosh x}}} dx \cr
& {\text{Integrate by }}\int {\frac{{du}}{u}} = \ln \left| u \right| + C,{\text{ then}} \cr
& \int {\frac{{\sinh x}}{{\cosh x}}} dx = \ln \left| {\cosh x} \right| + C \cr
& {\text{Therefore,}} \cr
& \int_0^{\ln 2} {\tanh x} dx = \left[ {\ln \left| {\cosh x} \right|} \right]_0^{\ln 2} \cr
& {\text{Evaluating}} \cr
& = \ln \left| {\cosh \left( {\ln 2} \right)} \right| - \ln \left| {\cosh \left( 0 \right)} \right| \cr
& = \ln \left| {\frac{5}{4}} \right| - \ln \left| 1 \right| \cr
& = \ln \left( {\frac{5}{4}} \right) \cr} $$
