Answer
$$\frac{\pi }{{12}}$$
Work Step by Step
$$\eqalign{
& \int_0^{\ln \sqrt 3 } {\frac{{{e^x}dx}}{{1 + {e^{2x}}}}} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = {e^x},{\text{ so that }}du = {e^x}dx \cr
& {\text{the new limits on }}t{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = \ln \sqrt 3,{\text{ then }}u = {e^{\ln \sqrt 3 }} = \sqrt 3 \cr
& \,\,\,\,\,\,{\text{If }}x = 0,{\text{ then }}u = {e^0} = 1 \cr
& {\text{then}} \cr
& \int_0^{\ln \sqrt 3 } {\frac{{{e^x}dx}}{{1 + {e^{2x}}}}} = \int_1^{\sqrt 3 } {\frac{{du}}{{1 + {u^2}}}} \cr
& {\text{integrate by using the formula }}\int {\frac{{du}}{{{a^2} + {u^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 1 \cr
& = \left( {{{\tan }^{ - 1}}\left( u \right)} \right)_1^{\sqrt 3 } + C \cr
& = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) - {\tan ^{ - 1}}\left( 1 \right) \cr
& = \frac{\pi }{3} - \frac{\pi }{4} \cr
& = \frac{\pi }{{12}} \cr} $$
