Answer
$\arctan \left( e \right) - \frac{\pi }{4}$
Work Step by Step
$$\eqalign{
&\text{Let }I= \int_0^1 {\frac{{{e^x}}}{{1 + {e^{2x}}}}} dx \cr
& = \int_0^1 {\frac{{{e^x}}}{{1 + {{\left( {{e^x}} \right)}^2}}}} dx \cr
& {\text{Let }}u = {e^x},{\text{ }}du = {e^x}dx \cr
& {\text{The new limits of integration are:}} \cr
& x = 1 \to u = {e^1},{\text{ }}u = e \cr
& x = 0 \to u = {e^0},{\text{ }}u = 1 \cr
& {\text{Applying the substitution}} \cr
& I = \int_1^e {\frac{{du}}{{1 + {u^2}}}} \cr
& {\text{Integrate }} \cr
& I= \left[ {\arctan u} \right]_1^e \cr
& {\text{Evaluating the limits of integration}} \cr
& I = \left[ {\arctan \left( e \right) - \arctan \left( 1 \right)} \right] \cr
& = \arctan \left( e \right) - \frac{\pi }{4} \approx 0.4328 \cr} $$
