Answer
\begin{aligned}
\int p^{4} e^{-p} d p =-(p^4 +4p^3 +12p^2 +24 +24)e^{-p}+C
\end{aligned}
Work Step by Step
Given $$\int p^{4} e^{-p} d p $$
So, we have
\begin{array}{|c c c|}\hline Differentiation && {Integration} \\
\hline
p^4 & + & e^{-p} \\
&\searrow&\\ \hline
4p^3 & - & -e^{-p} \\
&\searrow&\\ \hline
12p^2 &+ & e^{-p} \\
&\searrow&\\ \hline
24p & - &- e^{-p} \\
&\searrow&\\ \hline
24 &+ & e^{-p} \\
&\searrow&\\ \hline
0 & & -e^{-p} \\
&&\\ \hline
\end{array}
Therefore
\begin{aligned}
I&=\int p^{4} e^{-p} d p\\
&=-p^4e^{-p}-4p^3e^{-p}-12p^2e^{-p}-24e^{-p}-24e^{-p}+C\\
&=-(p^4 +4p^3 +12p^2 +24 +24)e^{-p}+C\\
\end{aligned}
