Answer
$$\frac{1}{{2\ln 2}}\ln \left( {1 + {2^{{x^2}}}} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{x{2^{{x^2}}}}}{{1 + {2^{{x^2}}}}}} dx \cr
& {\text{Use substitution}}\cr
& {\text{Let }}u = 1 + {2^{{x^2}}}\cr
& {\text{Then }}du = {2^{{x^2}}}\ln 2\left( {2x} \right)dx,\,\,\,\,dx = \frac{{du}}{{2x\left( {{2^{{x^2}}}} \right)\ln 2}} \cr
& {\text{write the integral in terms of }}u \cr
& \int {\frac{{x{2^{{x^2}}}}}{{1 + {2^{{x^2}}}}}} dx = \int {\frac{{x{2^{{x^2}}}}}{u}} \frac{{du}}{{2x\left( {{2^{{x^2}}}} \right)\ln 2}} \cr
& {\text{cancel the common factors}} \cr
& = \int {\frac{1}{u}} \frac{{du}}{{2\ln 2}} \cr
& = \frac{1}{{2\ln 2}}\int {\frac{1}{u}} du \cr
& {\text{integrate}} \cr
& = \frac{1}{{2\ln 2}}\ln \left| u \right| + C \cr
& {\text{write in terms of }}x{\text{; replace }}1 + {2^{{x^2}}}{\text{ for }}u \cr
& = \frac{1}{{2\ln 2}}\ln \left( {1 + {2^{{x^2}}}} \right) + C \cr} $$
