<s>[INST] 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?
 [/INST] 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}
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