Answer
$$\frac{1}{2} x^2 e^{x^2}-\frac{1}{2} e^{x^2}+C$$
Work Step by Step
Given $$\int x^3e^{x^2} d x $$
Let $u=x^{2}\ \ \ \ \ du = 2xdx$, then
\begin{aligned}
\int x^3e^{x^2} d x&=\frac{1}{2}\int ue^{u} du
\end{aligned}
Let \begin{align*}
u&=u \ \ \ \ \ \ \ dv=e^udu\\
du&=du\ \ \ \ \ \ \ v =e^u
\end{align*}
Then
\begin{aligned} \int x^3e^{x^2} d x&=\frac{1}{2}\int ue^{u} du\\
&= \frac{1}{2}\left[u \cdot e^{u}-\int e^{u} d u\right] \\ &=\frac{1}{2} u e^{u}-\frac{1}{2} e^{u}+C \\
&=\frac{1}{2} x^2 e^{x^2}-\frac{1}{2} e^{x^2}+C
\end{aligned}
