Answer
$$\frac{1}{3}x{e^{3x}} - \frac{1}{9}{e^{3x}} + C$$
Work Step by Step
$$\eqalign{
& \int {x{e^{3x}}} dx \cr
& {\text{substitute }}u = x,{\text{ }}du = dx \cr
& dv = {e^{3x}}dx,{\text{ }}v = \frac{1}{3}{e^{3x}} \cr
& {\text{ integration by parts}} \cr
& \int {udv} = uv - \int {vdu} \cr
& {\text{, we have}} \cr
& \int {x{e^{3x}}} dx = \frac{1}{3}x{e^{3x}} - \int {\left( {\frac{1}{3}{e^{3x}}} \right)dx} \cr
& \int {x{e^{3x}}} dx = \frac{1}{3}x{e^{3x}} - \frac{1}{3}\int {{e^{3x}}dx} \cr
& {\text{find antiderivative}} \cr
& \int {x{e^{3x}}} dx = \frac{1}{3}x{e^{3x}} - \frac{1}{3}\left( {\frac{1}{3}{e^{3x}}} \right) + C \cr
& \int {x{e^{3x}}} dx = \frac{1}{3}x{e^{3x}} - \frac{1}{9}{e^{3x}} + C \cr} $$
