Answer
$$\frac{\pi }{6} - {\sin ^{ - 1}}\left( {\frac{1}{4}} \right)$$
Work Step by Step
$$\eqalign{
& \int_{\ln 2}^{\ln 4} {\frac{{{e^{ - x}}}}{{\sqrt {1 - {e^{ - 2x}}} }}} dx \cr
& {\text{Let }}u = {e^{ - x}},{\text{ }}du = - {e^{ - x}}dx,{\text{ }}dx = \frac{1}{{ - {e^{ - x}}}}du \cr
& {\text{The new limits of integration are}} \cr
& x = \ln 4,{\text{ }}u = {e^{ - \ln 4}} = \frac{1}{4} \cr
& x = \ln 2,{\text{ }}u = {e^{ - \ln 2}} = \frac{1}{2} \cr
& {\text{Substituting}} \cr
& \int_{\ln 2}^{\ln 4} {\frac{{{e^{ - x}}}}{{\sqrt {1 - {e^{ - 2x}}} }}} dx = \int_{1/2}^{1/4} {\frac{{{e^{ - x}}}}{{\sqrt {1 - {u^2}} }}} \left( {\frac{1}{{ - {e^{ - x}}}}} \right)du \cr
& = - \int_{1/2}^{1/4} {\frac{1}{{\sqrt {1 - {u^2}} }}} du \cr
& {\text{Use the formula }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = {{\sin }^{ - 1}}u + C} \cr
& = - \left[ {{{\sin }^{ - 1}}u} \right]_{1/2}^{1/4} \cr
& = - \left[ {{{\sin }^{ - 1}}\left( {\frac{1}{4}} \right) - {{\sin }^{ - 1}}\left( {\frac{1}{2}} \right)} \right] \cr
& = - \left[ {{{\sin }^{ - 1}}\left( {\frac{1}{4}} \right) - \frac{\pi }{6}} \right] \cr
& = \frac{\pi }{6} - {\sin ^{ - 1}}\left( {\frac{1}{4}} \right) \cr} $$