I know it uses convolution theorem of inverse Laplace function but wasn't able to apply it . 
Find inverse laplace for:
     $$f(s)=\dfrac  1{(s-2)(s+2)^2}  $$

I know it uses convolution theorem of inverse Laplace function but wasn't able to apply it .
 A: If you have to use the Convolution Theorem then :
$$f(s)=\dfrac 1 {(s-2)(s+2)^2}$$
$$f(t)= \int_0^t e^{2(t-\tau)}\tau e^{-2 \tau} d\tau $$
$$f(t)= e^{2t}\int_0^t e^{-4\tau}\tau  d\tau $$
$$f(t)=\frac 1 {16}e^{2t}-\frac 1 {16}e^{-2t}-\frac 14 te^{-2t}$$

A simpler way:
$$f(s)=\dfrac 1 {(s-2)(s+2)^2}$$
Decompose into simple fractions:
$$f(s)=\dfrac A {(s-2)}+\dfrac B {(s+2)}+\dfrac C {(s+2)^2}$$
You should find:
$$(A,B,C)=\left (\frac 1 {16},-\frac 1 {16},-\frac 14 \right)$$
Then apply inverse Laplace Transform on each fraction.
$$f(t)=\frac 1 {16}e^{2t}-\frac 1 {16}e^{-2t}-\frac 14 te^{-2t}$$
A: The inverse Laplace Transform is given by
$$f(t)=\frac1{2\pi i}\int_{2^+-i\infty}^{2^+ +i\infty}\frac{1}{(s-2)(s+2)^2}e^{st}\,ds$$
For $t>0$, we close the contour in the left-half plane $s<2^+$.  Note that $F(s)e^{st}$ has a simple pole at $s=2$ and a second-order pole at $s=-2$.  
The associated residues of $F(s)e^{st}$ are $\frac{e^{2t}}{16}$ at $s=2$ and  $-\frac14 te^{-2t}-\frac1{16}e^{-2t}$ at $s=-2$.
Hence, we find that for $t>0$
$$f(t)=\frac1{16}e^{2t}-\frac1{16}e^{-2t}-\frac14 te^{-2t}$$

For $t<0$, we close the contour in the right-half plane.  Since $F(s)e^{st}$ is analytic in that plane, $f(t)=0$ for $t<0$.
