Answer
\begin{aligned}
\int x^{3} e^{x} d x
=(x^3 -3x^2 +6x -6)e^{x}+C
\end{aligned}
Work Step by Step
Given $$\int x^{3} e^{x} d x $$
So, we have
\begin{array}{|c c c|}\hline Differentiation && {Integration} \\
\hline
x^3 & + & e^x \\
&\searrow&\\ \hline
3x^2 & - & e^x \\
&\searrow&\\ \hline
6x &+ & e^x \\
&\searrow&\\ \hline
6 & - & e^x \\
&\searrow&\\ \hline
0 & & e^x \\
&&\\ \hline
\end{array}
Therefore
\begin{aligned}
I&=\int x^{3} e^{x} d x\\
&=x^3e^{x}-3x^2e^{x}+6xe^{x}-6e^{x}+C\\
&=(x^3 -3x^2 +6x -6)e^{x}+C
\end{aligned}
