Answer
\[\frac{1}{2}e^{x^2}\left[x^2 -1\right]+C\]
Where $C$ is constant of integration
Work Step by Step
Let \[I=\int x^3 e^{x^2}dx\]
\[I=\int x(x^2 e^{x^2})dx\]
Let $t=x^2 $ ___(1)
$\Rightarrow dt=2xdx$
\[I=\frac{1}{2}\int te^{t}dt\]
Using integration by parts
\[I=\frac{1}{2}\left[t\int e^t dt-\int \left((t)'\int e^{t}dt\right)dt\right]\]
\[I=\frac{1}{2}\left[te^t -\int e^{t}dt\right]\]
\[I=\frac{1}{2}\left[te^t -e^{t}\right]+C\]
Where $C$ is constant of integration
Using (1)
\[I=\frac{1}{2}e^{x^2}\left[x^2 -1\right]+C\]
Hence,
\[I=\frac{1}{2}e^{x^2}\left[x^2 -1\right]+C.\]
