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