Answer
$$\frac{1}{2}\left( {1 - \ln \left( {1 + e} \right) + \ln 2} \right)$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{x}{{1 + {e^{{x^2}}}}}} dx \cr
& {\text{Let }}u = {x^2},{\text{ }}du = 2xdx,{\text{ }}dx = \frac{{du}}{{2x}} \cr
& {\text{The new limits of integration are}} \cr
& x = 1 \to u = {\left( 1 \right)^2} = 1 \cr
& x = 0 \to u = {\left( 0 \right)^2} = 0 \cr
& {\text{Substituting}} \cr
& \int_0^1 {\frac{x}{{1 + {e^{{x^2}}}}}} dx = \int_0^{\pi /4} {\frac{x}{{1 + {e^u}}}\left( {\frac{1}{{2x}}} \right)} du \cr
& = \frac{1}{2}\int_0^1 {\frac{1}{{1 + {e^u}}}} du \cr
& {\text{Integrate by tables }}\int {\frac{1}{{1 + {e^u}}}du = u - \ln \left( {1 + {e^u}} \right) + C,{\text{ then}}} \cr
& \frac{1}{2}\int_0^1 {\frac{1}{{1 + {e^u}}}} du = \frac{1}{2}\left[ {u - \ln \left( {1 + {e^u}} \right)} \right]_0^1 \cr
& {\text{Evaluate and simplify}} \cr
& = \frac{1}{2}\left[ {1 - \ln \left( {1 + {e^1}} \right)} \right] - \frac{1}{2}\left[ {0 - \ln \left( {1 + {e^0}} \right)} \right] \cr
& = \frac{1}{2}\left( {1 - \ln \left( {1 + {e^1}} \right)} \right) - \frac{1}{2}\left( {0 - \ln 2} \right) \cr
& = \frac{1}{2} - \frac{1}{2}\ln \left( {1 + e} \right) + \frac{1}{2}\ln 2 \cr
& = \frac{1}{2}\left( {1 - \ln \left( {1 + e} \right) + \ln 2} \right) \cr} $$
