Answer
$$ = ta{n^{ - 1}}\left( {{e^x}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^x}}}{{1 + {e^{2x}}}}} dx \cr
& or \cr
& \int {\frac{{{e^x}}}{{1 + {{\left( {{e^x}} \right)}^2}}}} dx \cr
& {\text{substitutue }}u = {e^x},{\text{ }}du = {e^x}dx, \cr
& \int {\frac{{{e^x}}}{{1 + {{\left( {{e^x}} \right)}^2}}}} dx = \int {\frac{{du}}{{1 + {u^2}}}} \cr
& {\text{ use the formula }}\int {\frac{{du}}{{{a^2} + {u^2}}}} = \frac{1}{a}ta{n^{ - 1}}\left( {\frac{u}{a}} \right) + C,{\text{ }}\left( {{\text{see page 468}}} \right) \cr
& = \frac{1}{1}ta{n^{ - 1}}\left( {\frac{u}{1}} \right) + C \cr
& = ta{n^{ - 1}}\left( u \right) + C \cr
& {\text{write in terms of }}x \cr
& = ta{n^{ - 1}}\left( {{e^x}} \right) + C \cr} $$
