Answer
$$x^{4}e^x-4x^{3}e^x+12x^{2}e^x-24 \left[x e^x- e^x \right]+c$$
Work Step by Step
Given
$$ \int x^{4}e^x d x$$
by using the form
$$ \int x^{n} e^xd x = x^{n}e^x-n \int x^{n-1} e^xd x $$
\begin{align*}
\int x^{4}e^x d x&=x^{4}e^x-4 \int x^{3} e^xd x\\
&=x^{4}e^x-4\left[x^{3}e^x-3 \int x^{2} e^xd x\right] \\
&=x^{4}e^x-4x^{3}e^x+12\left[x^{2}e^x-2 \int x e^xd x\right] \\
&=x^{4}e^x-4x^{3}e^x+12x^{2}e^x-24 \left[x e^x- e^x \right]+c
\end{align*}
