Answer
$${\tan ^{ - 1}}\left( 5 \right) - \frac{\pi }{4}$$
Work Step by Step
$$\eqalign{
& \int_0^{\ln 5} {\frac{{{e^x}}}{{1 + {e^{2x}}}}} dx \cr
& {\text{Let }}u = {e^x},{\text{ }}du = {e^x}dx,{\text{ }}dx = \frac{1}{{{e^x}}}du \cr
& {\text{The new limits of integration are}} \cr
& x = \ln 5,{\text{ }}u = {e^{\ln 5}} = 5 \cr
& x = 0,{\text{ }}u = {e^0} = 1 \cr
& {\text{Substituting}} \cr
& \int_0^{\ln 5} {\frac{{{e^x}}}{{1 + {e^{2x}}}}} dx = \int_1^5 {\frac{{{e^x}}}{{1 + {u^2}}}} \left( {\frac{1}{{{e^x}}}} \right)du \cr
& = \int_1^5 {\frac{1}{{1 + {u^2}}}} du \cr
& {\text{Integrating}} \cr
& = \left[ {{{\tan }^{ - 1}}u} \right]_1^5 \cr
& = {\tan ^{ - 1}}\left( 5 \right) - {\tan ^{ - 1}}\left( 1 \right) \cr
& = {\tan ^{ - 1}}\left( 5 \right) - \frac{\pi }{4} \cr} $$
