Answer
$$\frac{1}{6}{e^{3{x^2}}} + C$$
Work Step by Step
$$\eqalign{
& \int {x{e^{3{x^2}}}} dx \cr
& {\text{set }}u = 3{x^2}{\text{ then }}\frac{{du}}{{dx}} = 6x,\,\,\,\,\,\,\frac{1}{6}du = xdx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {x{e^{3{x^2}}}} dx = \int {{e^u}} \left( {\frac{1}{6}du} \right) \cr
& = \frac{1}{6}\int {{e^u}} du \cr
& {\text{integrating by the formula }}\int {{e^u}} du = {e^u} + C \cr
& = \frac{1}{6}{e^u} + C \cr
& {\text{replace }}3{x^2}{\text{ for }}u \cr
& = \frac{1}{6}{e^{3{x^2}}} + C \cr} $$
