Answer
\begin{aligned}
\int t^2 e^{4t} d x
=( \frac{1}{4}t^2 - \frac{1}{8}t + \frac{1}{32} )e^{4t}+C \\
\end{aligned}
Work Step by Step
Given $$\int t^{2} e^{4t}d x $$
So, we have
\begin{array}{|c c c c|}\hline & Differentiation && {Integration} \\
\hline
+ \quad \rightarrow&t^{2} & & e^{4t} \\
&&\searrow&\\ \hline
- \quad \rightarrow& 2t & &\frac{1}{4}e^{4t} \\
&&\searrow&\\ \hline
+ \quad \rightarrow&2& &\frac{1}{16}e^{4t} \\
&&\searrow&\\ \hline
&0 & &\frac{1}{64}e^{4t} \\
&&\\ \hline
\end{array}
Therefore
\begin{aligned}
I&=\int t^2 e^{4t} d x\\
&= \frac{1}{4}t^2e^{4t} - \frac{1}{8}te^{4t} + \frac{1}{32} e^{4t}+C \\
&=( \frac{1}{4}t^2 - \frac{1}{8}t + \frac{1}{32} )e^{4t}+C \\
\end{aligned}
