I've gotten this far and I cannot proceed:
$L[y]=\frac{L[3e^{-x}sin(x)]+3}{p^2+2p+5}= \frac{3}{((p+1)^2+1)(p^2+2p+5)}+\frac{3}{p^2+2p+5}$
I'm finding it impossible to find the inverse to solve for $y$. I would guess partial fraction decomposition is next, but I cannot do it for these quantities. Help?
It's easier than it looks. Just write out $$\frac3{(p^p+2p+5)(p^2+2p+2)}=\frac{Ap+B}{p^2+2p+5}+\frac{Cp+D}{p^2+2p+2}$$ Multiply out that denominator $$3=(Ap+B)(p^2+2p+2)+(Cp+D)(p^2+2p+5))$$ Then compare coefficients of like powers of $p$ $$\begin{align}0 & =A+C&(1)\\ 0 &=2A+B+2C+D&(2)\\ 0 &=2A+2B+5C+2D&(3)\\ 3 &=2B+5D&(4)\end{align}$$ Now subtract twice eq $(1)$ from eqs $(2)$ and $(5)$ $$\begin{align}0 &=B+D&(5)\\ 0&=2B+3C+2D&(6)\end{align}$$ Subtract twice eq $(5)$ from eqs $(6)$ and $(4)$ $$\begin{align}0&=3C&(7)\\ 3&=3D&(8)\end{align}$$ And back into eq $(1)$ we have $$\begin{align}0&=A&(9)\end{align}$$ At this point from eq (8), we can see that $D=1$ and then from eq $(5)$, $B=-1$. So $$\frac3{(p^p+2p+5)(p^2+2p+2)}=\frac{-1}{p^2+2p+5}+\frac{1}{p^2+2p+2}$$ So I hope you can continue from this point.