Answer
$$ - \frac{\pi }{6}$$
Work Step by Step
$$\eqalign{
& \int_{\ln 2}^{\ln \left( {2/\sqrt 3 } \right)} {\frac{{{e^{ - x}}dx}}{{\sqrt {1 - {e^{ - 2x}}} }}} \cr
& or \cr
& \int_{\ln 2}^{\ln \left( {2/\sqrt 3 } \right)} {\frac{{{e^{ - x}}dx}}{{\sqrt {1 - {{\left( {{e^{ - x}}} \right)}^2}} }}} \cr
& {\text{substitute }}u = {e^{ - x}},{\text{ }}du = - {e^{ - x}}dx \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = \ln \left( {2/\sqrt 3 } \right),{\text{ }}u = {e^{ - \ln \left( {2/\sqrt 3 } \right)}} = \frac{{\sqrt 3 }}{2} \cr
& {\text{if }}x = \ln 2,{\text{ }}u = {e^{ - \ln 2}} = \frac{1}{2} \cr
& {\text{so}} \cr
& \int_{\ln 2}^{\ln \left( {2/\sqrt 3 } \right)} {\frac{{{e^{ - x}}dx}}{{\sqrt {1 - {{\left( {{e^{ - x}}} \right)}^2}} }}} = - \int_{1/2}^{\frac{{\sqrt 3 }}{2}} {\frac{1}{{\sqrt {1 - {u^2}} }}} du \cr
& {\text{find the antiderivative}} \cr
& = \left. {\left( {{{\cos }^{ - 1}}u} \right)} \right|_{1/2}^{\frac{{\sqrt 3 }}{2}} \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = {\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) - {\cos ^{ - 1}}\left( {\frac{1}{2}} \right) \cr
& {\text{simplify}} \cr
& = {\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) - {\cos ^{ - 1}}\left( {\frac{1}{2}} \right) \cr
& = \frac{\pi }{6} - \frac{\pi }{3} \cr
& = - \frac{\pi }{6} \cr} $$
