Answer
$$\displaystyle\int{x^5e^{-x^3}dx}=-\frac{1}{3}e^{-x^3}\left( x^3+1 \right) +C$$
Work Step by Step
$I=\displaystyle\int{x^5e^{-x^3}dx}$
$\displaystyle{\left[\begin{matrix}
u=-x^3& dx=-\frac{du}{3x^2}\\
\end{matrix}\right]}$
$I=\displaystyle\int{x^5e^u}\times -\frac{du}{3x^2}\\
I=\displaystyle\frac{1}{3}\int{-x^3e^u}du\\
I=\displaystyle\frac{1}{3}\int{ue^u}du$
$\displaystyle{\left[\begin{matrix}
u=u& du=1\\
dv=e^u& v=e^u\\
\end{matrix}\right]}$
$I=\displaystyle\frac{1}{3}\left( ue^u-\int{e^udu} \right) \\
I=\displaystyle\frac{1}{3}\left( ue^u-e^u \right) +C\\
I=\displaystyle\frac{1}{3}\left( -x^3e^{-x^3}-e^{-x^3} \right) +C\\
I=\displaystyle-\frac{1}{3}e^{-x^3}\left( x^3+1 \right) +C$
