Answer
$$ \dfrac{e^{-3 s}}{s^2}+ \dfrac{3 e^{-3 s}}{s}$$
Work Step by Step
We are given that $f(t)= t u_{3} (t)\\=(t-3) u_3(t) +3 u_3 (t) $
Now,$F(s)=L[(t-3) u_3(t) +3 u_3 (t)] \\= \dfrac{e^{-3 s}}{s^2}+ \dfrac{3 e^{-3 s}}{s}\\= \dfrac{e^{-3 s}}{s^2}+ \dfrac{3 e^{-3 s}}{s}$
