Answer
$$\int te^{-3t}dt=-\frac{1}{3}te^{-3t}-\frac{1}{9}e^{-3t}+C$$
Work Step by Step
$$A=\int te^{-3t}dt$$
We would choose $u=t$ and $dv=e^{-3t}dt$
We could easily see that $du=dt$.
For $dv=e^{-3t}dy$, we analyze as follows $$\int e^{-3t}dt=\int\frac{1}{-3}e^{-3t}d(-3t)=-\frac{1}{3}\int e^{-3t}d(-3t)=-\frac{1}{3}e^{-3t}+C$$
Therefore, for $dv=e^{-3t}dt$, $v=-\frac{1}{3}e^{-3t}$
Apply Integration by Parts to A, we have $$A=uv-\int vdu$$ $$A=-\frac{1}{3}te^{-3t}-\int(-\frac{1}{3}e^{-3t})dt$$ $$A=-\frac{1}{3}te^{-3t}+\frac{1}{3}\int e^{-3t}dt$$ $$A=-\frac{1}{3}te^{-3t}+\frac{1}{3}(-\frac{1}{3}e^{-3t}+C)$$ $$A=-\frac{1}{3}te^{-3t}-\frac{1}{9}e^{-3t}+C$$
