Answer
$$2 - \frac{\pi }{2}$$
Work Step by Step
$$\eqalign{
& \int_0^{\ln 2} {\sqrt {{e^x} - 1} } dx \cr
& {\text{Integrate by the substitution method}} \cr
& {u^2} = {e^x} - 1,\,\,\,\,\,\,\,\,\,\,\,2udu = {e^x}dx,\,\,\,\,\,\,\,\,\,\,dx = \frac{{2udu}}{{{e^x}}} = \frac{{2udu}}{{{u^2} + 1}} \cr
& \,\,\,\,x = \ln 2 \to u = 1\,\,\,\,\,\,\,\,\,\,{\text{ and}}\,\,\,\,\,\,\,\,\,{\text{ }}x = 0 \to \,\,u = 0 \cr
& \cr
& {\text{write the integral in terms of }}u \cr
& \int_0^{\ln 2} {\sqrt {{e^x} - 1} } dx = \int_0^1 {\sqrt {{u^2}} } \left( {\frac{{2udu}}{{{u^2} + 1}}} \right) \cr
& = \int_0^1 {\frac{{2{u^2}du}}{{{u^2} + 1}}} \cr
& {\text{Use long division}} \cr
& = \int_0^1 {\left( {2 - \frac{2}{{{u^2} + 1}}} \right)} du \cr
& \cr
& {\text{integrate and simplify}} \cr
& = \left[ {2u - 2{{\tan }^{ - 1}}\left( u \right)} \right]_0^1 \cr
& = 2\left[ {u - {{\tan }^{ - 1}}\left( u \right)} \right]_0^1 \cr
& = 2\left[ {1 - {{\tan }^{ - 1}}\left( 1 \right)} \right] - 2\left[ {0 - {{\tan }^{ - 1}}\left( 0 \right)} \right] \cr
& = 2\left( {1 - \left( {\frac{\pi }{4}} \right)} \right) \cr
& = 2 - \frac{\pi }{2} \cr} $$
