Answer
$$\frac{1}{2}{x^2}{e^{{x^2}}} - \frac{1}{2}{e^{{x^2}}} + C$$
Work Step by Step
$$\eqalign{
& \int {{x^3}{e^{{x^2}}}} dx \cr
& {\text{substitute }}u = {x^2},{\text{ }}du = 2xdx \cr
& dv = x{e^{{x^2}}}dx,{\text{ }} \cr
& v = \int {x{e^{{x^2}}}dx} \cr
& v = \frac{1}{2}{e^{{x^2}}} \cr
& {\text{ integration by parts}} \cr
& \int {udv} = uv - \int {vdu} \cr
& {\text{, we have}} \cr
& = \frac{1}{2}{x^2}{e^{{x^2}}} - \int {\left( {\frac{1}{2}{e^{{x^2}}}} \right)\left( {2x} \right)dx} \cr
& = \frac{1}{2}{x^2}{e^{{x^2}}} - \int {x{e^{{x^2}}}dx} \cr
& {\text{find antiderivative}} \cr
& = \frac{1}{2}{x^2}{e^{{x^2}}} - \frac{1}{2}{e^{{x^2}}} + C \cr} $$
