Answer
$\frac{xe^{3x}}{3}-\frac{e^{3x}}{9}+C$
Work Step by Step
We use the formula
$\int udv=uv-\int vdu$
with $u=x$, $dv=e^{3x}dx$
$du=dx$, $v=\frac{e^{3x}}{3}$
Then $\int xe^{3x}dx=x\times \frac{e^{3x}}{3}-\int \frac{e^{3x}}{3}dx$
$=\frac{xe^{3x}}{3}-\frac{e^{3x}}{9}+C$
