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.

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
$$
\mathcal{L}[e^{at}\cos bt] = \frac{s-a}{(s-a)^2 + b^2}
$$
