Using Laplace transform to solve an IVP The original problem is: 
$y''+9y = e^{-t} \sin (t)$
and the initial conditions are: $y(0)=-1$ and $y'(0)=1$.
I've taken the laplace of both sides and got:
$Y(s^2+9) = \frac{1}{(s+1)^2 +1} -s +1$
Once I divide through by $s^2 + 9$ the problems begin. The last two terms are straightforward but I'm having trouble finding an inverse laplace for the first term which is $\frac{1}{(s^2+9)(s^2+2s+2)}$.
I've tried to decompose it into the addition of two fractions by partial fraction decomposition by I keep failing to find two terms for A and B. What should i try or have i made a mistake so far in the steps i've taken?

Everything up to this point looks good:
$$Y(s) = \frac{1}{(s^2+9)((s+1)^2+1)} - s + 1$$
To do partial fractions on $\dfrac{1}{(s^2+9)((s+1)^2+1)}$, note that
$$\frac{1}{(s^2+9)(s^2+2s+2)} = \frac{As+B}{s^2+9} + \frac{Cs+D}{s^2+2s+2}$$ since each of $s^2+9$ and $s^2+2s+2$ are irreducible quadratics.
I leave it to you to show that

 $$\dfrac{1}{(s^2+9)(s^2+2s+2)} = -\dfrac{2s+7}{85(s^2+9)} + \dfrac{2s+11}{85(s^2+2s+2)}$$


EDIT: Alternatively, you could note that
$$\mathcal{L}^{-1}\left\{\frac{1}{(s^2+9)((s+1)^2+1)}\right\} = \mathcal{L}^{-1}\left\{\frac{1}{s^2+9}\right\}\ast\mathcal{L}^{-1}\left\{\frac{1}{(s+1)^2+1}\right\} = \frac{1}{3}\sin(3t)\ast e^{-t}\sin t$$
where $f(t)\ast g(t) = \displaystyle\int_0^t f(t-\tau)g(\tau)\,d\tau$ denotes convolution.  The computation of the convolution can be found in the spoiler below.

$$\begin{aligned}&\phantom{=} \frac{1}{3}\sin(3t)\ast e^{-t}\sin t\\ &= \frac{1}{3}\int_0^t \sin(3t-3\tau)e^{-\tau}\sin\tau \,d\tau\\ &= \frac{1}{6}\int_0^te^{-\tau}(\cos(3t-3\tau-\tau)-\cos(3t-3\tau+\tau))\,d\tau\\ &= \frac{1}{6}\left[\int_0^te^{-\tau}\cos(3t-4\tau)\,d\tau -\int_0^t e^{-\tau}\cos(3t-2\tau)\,d\tau\right]\\ &= \frac{1}{6}\left[\left.\left[\frac{e^{-\tau}}{17}(-4\sin(3t-4\tau)-\cos(3t-4\tau)\right]\right|_0^t - \left.\left[\frac{e^{-\tau}}{5}(-2\sin(3t-2\tau)-\cos(3t-2\tau)\right]\right|_0^t\right]\\ &= \frac{1}{6}\left[\frac{1}{17}\left[e^{-t}(4\sin t-\cos t)+4\sin(3t)+\cos(3t)\right]\right. \\ &\phantom{= \frac{1}{6}\left[\frac{1}{17}\left[e^{-t}(4\sin t-\cos t)\right]\right.}\left.+\frac{1}{5}\left[e^{-t}(2\sin t+\cos t)-2\sin(3t)-\cos(3t)\right]\right] \\ &=\frac{1}{6}\left[\frac{1}{85}\left[e^{-t}(54\sin t+12\cos t) - 14\sin(3t)-12\cos(3t)\right]\right]\\ &= \frac{1}{255}\left[e^{-t}(27\sin t+6\cos t) - 7\sin(3t)-6\cos(3t)\right]\end{aligned}$$

Where
$$\int e^{ax}\cos(bx+c)\,dx = \frac{e^{ax}}{a^2+b^2}(a\cos(bx+c)+b\sin(bx+c))+C$$
can easily be shown using integration by parts.
