Answer
$=\frac{1}{9}e^{3x}(3x-1)+C$
Work Step by Step
$\int xe^{3x}dx$
Use integration by parts once.
$u=x, dv=\int e^{3x}dx$
$u'=dx, v=\frac{1}{3}e^{3x}$
$=uv-\int u'vdx$
$=x\frac{1}{3}e^{3x}-\int \frac{1}{3}e^{3x}dx$
$=\frac{1}{3}xe^{3x}-\frac{1}{9}e^{3x}+C$
$=\frac{1}{9}e^{3x}(3x-1)+C$