Answer
$\frac{2}{s}+\frac{2e^{-s-1}}{s+1}-\frac{2e^{-s}}{s}$
Work Step by Step
Given: $f(t)=2+2(e^{-t}-1)u_1(t)$
Rewrite as:
$f(t)=2+2e^{-t}u_1(t)-2u_1(t)$
$L[2e^{-t}u_1(t)=e^{-s}L[F(t)]$
With $F(t)=2e^{-t}$
we have:
$L[f(s)]=2e^{-s}L[e^{-t}]\\
=\frac{2e^{-s-1}}{s+1}$
Using the Convolution Theorem
$L[f(s)]=L[2+2(e^{-t}-1)u_1(t)]\\
=\frac{2}{s}+\frac{2e^{-s-1}}{s+1}-\frac{2e^{-s}}{s}$
