Answer
$$\frac{1}{{2\ln 3}}\left( {{3^{{x^2}}}} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {x{3^{{x^2}}}} dx \cr
& {\text{use substitution}}{\text{: }} \cr
& {\text{ }}u = {x^2},{\text{ then }}\frac{{du}}{{dx}} = 2x,\,\,\,\,\frac{{du}}{{2x}} = dx \cr
& {\text{then}} \cr
& \int {x{3^{{x^2}}}} dx = \int {x{3^u}} \left( {\frac{{du}}{{2x}}} \right) \cr
& = \int {{3^u}} \left( {\frac{{du}}{2}} \right) \cr
& = \frac{1}{2}\int {{3^u}} du \cr
& {\text{integrate using }}\int {{a^u}} du = \frac{{{a^u}}}{{\ln a}} + C \cr
& = \frac{1}{2}\left( {\frac{{{3^u}}}{{\ln 3}}} \right) + C \cr
& {\text{replace }}{x^2}{\text{ for }}u \cr
& = \frac{1}{{2\ln 3}}\left( {{3^{{x^2}}}} \right) + C \cr} $$
