Answer
$$\frac{5}{32} e^{4}-\frac{1}{32}$$
Work Step by Step
Given $$\int_{0}^{1} x^{2} e^{4 x} d x$$
Let
\begin{align*}
u&= x^2\ \ \ \ \ \ \ \ \ dv= e^{4x}dx\\
du&= 2xdx\ \ \ \ \ \ \ \ \ v=\frac{1}{4} e^{4x}
\end{align*}
Then
\begin{align*}
\int e^{4 x} d x&= \frac{1}{4} x^{2} e^{4 x}-\frac{1}{2} \int x e^{4 x} d x
\end{align*}
\begin{align*}
u&= x \ \ \ \ \ \ \ \ \ dv= e^{4x}dx\\
du&= dx\ \ \ \ \ \ \ \ \ v=\frac{1}{4} e^{4x}
\end{align*}
Then
\begin{align*}
\int x e^{4 x} d x&=\frac{1}{4} x e^{4 x}-\frac{1}{4} \int e^{4 x} d x\\
&= \frac{1}{4} x e^{4 x}- \frac{e^{4 x}}{16}
\end{align*}
Hence
\begin{align*}
\int_{0}^{1} x^{2} e^{4 x} d x&= \frac{1}{4} x^{2} e^{4 x}-\frac{1}{8} x e^{4 x}+\frac{1}{32} e^{4 x}\bigg|_{0}^{1}\\
&=\frac{5}{32} e^{4}-\frac{1}{32}
\end{align*}
