Find the inverse Laplace of $ \ \frac{2}{(s-1)^3(s-2)^2}$. Find the inverse Laplace of  $ \ \frac{2}{(s-1)^3(s-2)^2}$. 
Answer:
To do this we have to make partial fractions as follows:
$ \frac{2}{(s-1)^3(s-2)^2}=\frac{A}{S-1}+\frac{B}{(s-1)^2}+\frac{C}{(s-1)}+\frac{J}{(s-1)^3}+\frac{K}{(s-2)^2}+\frac{L}{s-2}$
Am I right so far?
Does there is any easy way?
 A: You have one too many $\frac {1}{(s-1)}$ terms.
$\frac {2}{(s-1)^3(s-2)^2} = \frac {A}{(s-1)} + \frac {B}{(s-1)^2} + \frac {C}{(s-1)^3} + \frac {D}{(s-2)} + \frac {E}{(s-2)^2}$
Traditionally, you would then multiply through to clear out the denominators.
$2 = A(s-1)^2(s-2)^2 + B(s-1)(s-2)^2 + C(s-2)^2 + D(s-1)^3(s-2) + E(s-1)^3$
Here is a trick though.
$\lim_\limits{s\to 1} \frac {2(s-1)^3}{(s-1)^3(s-2)^2} =  C $
and 
$\lim_\limits{s\to 2} \frac {2(s-2)^2}{(s-1)^3(s-2)^2} =  E$
But, you might find it easier to do this in pieces.
If you start with 
$\frac {2}{(s-1)^3(s-2)^2} = \frac {As^2+Bs+C}{(s-1)^3} + \frac {Ds+E}{(s-2)^2}$
(Note that this will not have the same values for $A,B,C,D,E$ as the first approach)
Then there is less multiplying up front.  But, you might need to do more decomposition down the line.
A: Another way is using convolution
\begin{align}
\frac{2}{(s-1)^3(s-2)^2}
&= \frac{2}{(s-1)^3}\cdot\frac{1}{(s-2)^2} \\
&= {\cal L}\left(t^2e^t\right){\cal L}\left(te^{2t}\right) \\
&= \int_0^xt^2e^t(x-t)e^{2x-2t}\ dt \\
&= e^{2x}\int_0^xe^{-t}(xt^2-t^3)\ dt \\
&= \color{blue}{2e^{2x}(x-3)+e^x(x^2+4x+6)}
\end{align}
