Answer
$$2 + \frac{1}{2}\ln \left( {\frac{2}{{1 + {e^4}}}} \right)$$
Work Step by Step
$$\eqalign{
& y = \frac{x}{{1 + {e^{{x^2}}}}},{\text{ }}y = 0,{\text{ }}x = 2 \cr
& {\text{The area of the region is given by}} \cr
& A = \int_0^2 {\frac{x}{{1 + {e^{{x^2}}}}}} dx \cr
& {\text{Let }}t = {x^2},{\text{ }}dt = 2xdx,{\text{ }}dx = \frac{1}{{2x}}dt \cr
& {\text{The new limits of integration are:}} \cr
& x = 2 \to t = 4 \cr
& x = 0 \to t = 0 \cr
& A = \int_0^2 {\frac{x}{{1 + {e^{{x^2}}}}}} dx = \int_0^4 {\frac{x}{{1 + {e^t}}}\left( {\frac{1}{{2x}}} \right)} dt \cr
& = \frac{1}{2}\int_0^4 {\frac{1}{{1 + {e^t}}}} dt \cr
& {\text{Integrate by tables }}\int {\frac{1}{{1 + {e^u}}}du} = u - \ln \left( {1 + {e^u}} \right) + C \cr
& \frac{1}{2}\int_0^4 {\frac{1}{{1 + {e^t}}}} dt = \frac{1}{2}\left[ {t - \ln \left( {1 + {e^t}} \right)} \right]_0^4 \cr
& \left[ {t - \ln \left( {1 + {e^t}} \right)} \right]_0^4 = \frac{1}{2}\left[ {4 - \ln \left( {1 + {e^4}} \right)} \right] - \frac{1}{2}\left[ {0 - \ln \left( {1 + {e^0}} \right)} \right] \cr
& = \frac{1}{2}\left[ {4 - \ln \left( {1 + {e^4}} \right)} \right] - \frac{1}{2}\left[ { - \ln \left( 2 \right)} \right] \cr
& = 2 - \frac{1}{2}\ln \left( {1 + {e^4}} \right) + \frac{1}{2}\ln \left( 2 \right) \cr
& = 2 + \frac{1}{2}\ln \left( {\frac{2}{{1 + {e^4}}}} \right) \cr} $$
