Answer
$
=\frac{e^{-3s}(4s^2+5s+2)}{s^3}+\frac{s+1}{s^2}$
Work Step by Step
Obtain: $F(s)=\int^{\infty}_0 e^{-st}f(t)dt\\
=\int^3_0 e^{-st}(t+1)dt+\int^{\infty}_3 e^{-st}(t^2-1) dt\\
=\int^3_0 e^{-st}(t+1)dt+\lim \int^n_3 e^{-st}(t^2-1)dt\\
=[-\frac{t+1}{s}e^{-st}]^3_0+\frac{1}{s}\int^3_0 e^{-st}dt+\lim [(-\frac{t^2-1}{s}e^{-st})^n_3+\frac{2}{3}\int^n_3 te^{-st}dt)]\\
=(-\frac{4}{3}e^{-3s}+\frac{1}{s})+(-\frac{1}{s^2}e^{-3s}+\frac{1}{s^2})+(\frac{8}{s}e^{-3s}+\frac{6}{s^2}e^{-3s}+\frac{2}{s^3}e^{-3s})\\
=\frac{e^{-3s}(4s^2+5s+2)}{s^3}+\frac{s+1}{s^2}$
