Find inverse Laplace of $\displaystyle \frac{3s+2}{s^2+2s+10}$. Book's answer (using partial fractions) $\displaystyle 3^{-t}\cos 3t - \frac 13 e^{-t} \sin 3t$. My answer $\displaystyle 3^{-t}\cos 3t - \frac 23 e^{-t} \sin 3t$ (I have an extra 2 in the top of the second term).Where did my 2 go?

My work. Since my first term agrees with the book's answer, I'll just do inverse Laplace of

$\displaystyle \frac{2}{s^2+2s+10} = \frac{2}{(s+1)^2+9}.$

Now, to understand this I do it a little differently than is explained in the book (and various web pages).

If $\displaystyle F(s+1) = \frac{2}{(s+1)^2+9}$, then $\displaystyle F(s) = \frac{2}{(s)^2+9}$.

I want the inverse Laplace of F(s+1), but first I find the inverse L of F(s).

The Laplace of $\displaystyle \sin 3t = \frac 3{s^2+9}$. Multiply both sides by $\displaystyle \frac 23$ and I get that the inverse Laplace of $\displaystyle \frac 23 \sin 3t = \frac 2{s^2+9} = F(s)$. Thus the inverseL of $\displaystyle \frac 2{s^2+9}$ is $\displaystyle \frac 23 \sin 3t$.

But I want to find the inverse Laplace of $\displaystyle F(s+1)$ which gives me $\displaystyle e^{-t}\frac 23 \sin 3t$, and I still have an extra 2 in the numerator.What'd I do wrong?