Answer
$-\frac{1}{3} te^{-3t}-\frac{1}{9}e^{-3t}+C$
Work Step by Step
Let $u=t$ and $dv=e^{-3t} dt$
then $du = dt$ and $v=-\frac{1}{3}e^{-3t}$
and therefore
$\int te^{-3t}dt = -\frac{1}{3}te^{-3t}-\int -\frac{1}{3}e^{-3t}dt=-\frac{1}{3}te^{-3t}+\frac{1}{3}\int e^{-3t}dt=-\frac{1}{3} te^{-3t}-\frac{1}{9}e^{-3t}+C$
