Answer
$ 1-e^{-2t}-2 e^{-2t} t $
Work Step by Step
Since, $F(s)=\dfrac{4}{s(s+2)^2}=\dfrac{1}{s}-\dfrac{1}{s+2}-\dfrac{2}{(s+2)^2}$
The inverse Laplace transform of function can be expressed as:
$F(t)=L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+2}-\dfrac{2}{(s+2)^2}]$
Now, apply the first shifting Theorem.
$f(t)= 1-e^{-2t}-2 e^{-2t} t $
