Question

Evaluate the integral(integration by parts)
te^(-3t)dt

Answer 1

\[I=\int\limits t e ^{-3t}dt=t \frac{ e ^{-3t} }{-3 }-\int\limits 1*\frac{ e ^{-3t} }{-3 }dt+c\]
\[I=\frac{ te ^{-3t} }{-3 }+\frac{ e ^{-3t} }{ -3 }+c\]