## Calculus 10th Edition

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