Answer
$$ \frac{-(x+1)}{3}e^{4-3 x} -\frac{1}{9} e^{4-3 x}+C $$
Work Step by Step
Given $$\int(x+1) e^{4-3 x} d x $$
Let
\begin{align*}
u&= x+1\ \ \ \ \ \ dv=e^{4-3 x} d x\\
du&= dx\ \ \ \ \ \ v=\frac{-1}{3}e^{4-3 x}
\end{align*}
Then
\begin{align*}
\int(x+1) e^{4-3 x} d x&=\frac{-(x+1)}{3}e^{4-3 x} +\frac{1}{3}\int e^{4-3 x}dx\\
&= \frac{-(x+1)}{3}e^{4-3 x} -\frac{1}{9} e^{4-3 x}+C
\end{align*}
