Answer
$-\dfrac{1}{3} e^{-2t} +\dfrac{1}{3}e^{t}+e^t t$
Work Step by Step
Since, $F(s)=\dfrac{2s+1}{(s+2)(s-1)^2}=-\dfrac{1}{3(s+2)}+\dfrac{1}{3(s-1)}+\dfrac{1}{(s-1)^2}$
The inverse Laplace transform of function can be expressed as:
$F(t)=L^{-1} [-\dfrac{1}{3(s+2)}+\dfrac{1}{3(s-1)}+\dfrac{1}{(s-1)^2}]$
Now, apply the first shifting Theorem.
$f(t)= -\dfrac{1}{3} e^{-2t} +\dfrac{1}{3}e^{t}+e^t t$
