Answer
See below
Work Step by Step
The Laplace Transform can be written as:
$L[F(t)]=\int_{0}^{\infty} e^{-st} f(t) dt $
Now, $L[F(t)]=\int_{0}^{1} e^{-st} f(t) dt+\int_{1}^{3} e^{-st} f(t) dt+\int_{3}^{\infty} e^{-st} f(t) dt\\
=\int_{0}^{1} e^{-st} dt+\int_{1}^{3} e^{-st} dt+\int_{3}^{\infty} e^{-s(t-3)} dt\\
=\dfrac{1-e^{-s}-se^{-3s}}{s^2}+\frac{e^{-3s}}{s-1} $
