<s>[INST] Find the inverse laplace transform of the given problem Find the inverse laplace transform of $ \ Y(s)= \large \frac{\large \frac{82}{ \large s-6}-2s+2}{s^2+6s+10} \ $
Answer:
Let $ \ \mathcal{L}^{-1} \ $ be the inverse laplace operator.
Then,
$ y(t)=\mathcal{L}^{-1} [Y(s);t] \\ \Rightarrow y(t)= \mathcal{L}^{-1} \left[\frac{\large \frac{82}{ \large s-6}-2s+2}{s^2+6s+10}   \right] \ = \mathcal{L}^{-1} \left[\frac{ \large -2s^2+12s+70}{ \large (s-6)(s^2+6s+10)} \right] $
Now,
$ \frac{ \large -2s^2+12s+70}{ \large (s-6)(s^2+6s+10)}= \frac{A}{s-6}+\frac{\large Bs+C}{s^2+6s+10} \ $ where $ \ A,B,C \ $ are  unknown constants to be  determined.
Is this the correct partial fraction?
Help me find the inverse laplace transform.
 [/INST] You are on the right track, after you work out the constants $A$, $B$ and $C$ you should end up with
\begin{eqnarray}
Y(s) &=& \frac{1}{s - 6} - \frac{3s + 10}{s^2 + 6s + 10}  \\
&=& \frac{1}{s - 6} - \frac{3(s + 3) + 1}{(s + 3)^2 + 1} \\ 
&=& \frac{1}{s - 6} - 3\frac{(s + 3)}{(s + 3)^2 + 1} - \frac{1}{(s + 3)^2 + 1}
\end{eqnarray}
Now use the fact that
$$
\mathcal{L}[e^{at}\sin bt] = \frac{b}{(s-a)^2 + b^2}
$$
and
$$
\mathcal{L}[e^{at}\cos bt] = \frac{s-a}{(s-a)^2 + b^2}
$$
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