Please tell me if i did this inverse laplace correctly. Thanks The question is : find the inverse laplace transformation of 
$$\frac{13s^2+3s+6}{(s-2)(s^2+9)}.$$
Please tell me if i did this correctly
Here is my work:
Using partial factions:
\begin{align}
Y(s) &= \frac{13s^2 + 3s + 6}{(s - 2)(s^2 + 9)}
\\
&= \frac{64}{13 (-2 + s)} + \frac{3 (83 + 35 s)}{13 (9 + s^2)}
\end{align}
And: $$\frac{3 (83 + 35 s)}{13 (9 + s^2)} = \frac{83}{9 + s^2} + \frac{35s}{9 + s^2}$$
Finally, identity in partial fraction is:
\begin{equation}
Y(s) = \frac{64}{13 (-2 + s)} + \frac{83}{9 + s²} + \frac{35 s}{9 + s²}
\end{equation}
And then:
$Y(s) ↔ y(t)$, using the Laplace Transformation table.
The answer is:
$$y(t) = (64/13) e^{2t} u(t) + (83/3) \sin(3t) u(t) + 35\cos(3t) u(t)$$

The partial fraction expansion yields:
$\displaystyle 
\frac{3 (35 s+83)}{13 (s^2+9)} + \frac{64}{13 (s-2)} = \frac{3(35 s)}{13 (s^2+9)} + \frac{3(83)}{13 (s^2+9)}+ \frac{64}{13 (s-2)}$
Now, we put that result into the desired forms:
$\displaystyle \frac{3(35 s)}{13 (s^2+3^2)} + \frac{3(83)}{13 (s^2+3^2)}+ \frac{64}{13 (s-2)}$
From this, we can see the forms we need.
This yields:
$$\displaystyle y(t) = \frac{1}{13}\left(105 \cos 3t + 83 \sin 3t + 64 e^{2t}\right)$$
