Answer
$$ \int(4 x-12) e^{-8 x} d x =\frac{-1}{8}(4 x-12)e^{-8x}+\frac{-1}{16}e^{-8x}+c$$
Work Step by Step
Since $$\int(4 x-12) e^{-8 x} d x$$
Let
\begin{align*}
u&= (4 x-12)\ \ \ \ \ dv=e^{-8x}dx\\
du&= 4dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v=\frac{-1}{8}e^{-8x}
\end{align*}
Then integrate by parts , we get
\begin{align*}
\int(4 x-12) e^{-8 x} d x&=uv-\int vdu\\
&=\frac{-1}{8}(4 x-12)e^{-8x}+\frac{1}{2}\int e^{-8x}dx \\
&=\frac{-1}{8}(4 x-12)e^{-8x}+\frac{-1}{16}e^{-8x}+c
\end{align*}
