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}^{\infty} e^{-st} f(t) dt\\
=\int_{0}^{1} e^{-st} dt+\int_{1}^{\infty} e^{-st} dt\\
=\dfrac{te^{-s}}{-s}-\dfrac{e^{-s}}{s}+ \dfrac{1}{s^2}\\
=\dfrac{1-e^{-s}-se^{-s}}{s^2} $