University Calculus: Early Transcendentals (3rd Edition)

$$\int x e^{3x}dx=\frac{xe^{3x}}{3}-\frac{e^{3x}}{9}+C$$
$$A=\int x e^{3x}dx$$ Take $u=x$ and $dv=e^{3x}dx$ We then have $du=dx$ and $v=\frac{e^{3x}}{3}$ Apply the formula $\int udv= uv-\int vdu$, we have $$A=\frac{xe^{3x}}{3}-\int\frac{e^{3x}}{3}dx$$ $$A=\frac{xe^{3x}}{3}-\frac{1}{3}\int e^{3x}dx$$ $$A=\frac{xe^{3x}}{3}-\frac{1}{3}\times\frac{e^{3x}}{3}+C$$ $$A=\frac{xe^{3x}}{3}-\frac{e^{3x}}{9}+C$$