Answer
$$ \int(6 x+3) e^{-2 x} d x =\frac{-1}{2}(6 x+3)e^{-2x}-\frac{3}{2}e^{-2x}+c$$
Work Step by Step
Since $$\int(6 x+3) e^{-2 x} d x$$
Let
\begin{align*}
u&= (6x+3)\ \ \ \ \ dv=e^{-2x}dx\\
du&= 6dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v=\frac{-1}{2}e^{-2x}
\end{align*}
Then integrate by parts , we get
\begin{align*}
\int(6 x+3) e^{-2 x} d x&=uv-\int vdu\\
&=\frac{-1}{2}(6 x+3)e^{-2x}+3\int e^{-2x}dx \\
&=\frac{-1}{2}(6 x+3)e^{-2x}-\frac{3}{2}e^{-2x}+c
\end{align*}
