Answer
$$\ln \left( {\frac{5}{4}} \right)$$
Work Step by Step
$$\eqalign{
& \int_0^{\ln 2} {\tanh x} dx \cr
& {\text{using the hyperbolic identity }}\tanh x = \frac{{\sinh x}}{{\cosh x}} \cr
& = \int_0^{\ln 2} {\frac{{\sinh x}}{{\cosh x}}} dx \cr
& {\text{set }}u = \cosh x{\text{ }}\,\,\,\,\,{\text{then }}\,\,\,\,du = \sinh xdx, \cr
& {\text{switch the limits of integration}} \cr
& u = \cosh x,\,\,\,\,\,\,\,x = 0 \to u = \cosh 0 = 1 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \ln 2 \to u = \cosh \left( {\ln 2} \right) = \frac{{{e^{\ln 2}} + {e^{ - \ln 2}}}}{2} = \frac{5}{4} \cr
& {\text{use the change of variable}} \cr
& \int_1^{\ln 2} {\frac{{\sinh x}}{{\cosh x}}} dx = \int_1^{5/4} {\frac{{du}}{u}} \cr
& {\text{integrate and evaluate the limits}} \cr
& = \left( {\ln \left| u \right|} \right)_1^{5/4} \cr
& = \ln \left| {\frac{5}{4}} \right| - \ln \left| 1 \right| \cr
& = \ln \left( {\frac{5}{4}} \right) \cr} $$
