Computing the inverse Laplace transform of this?

Question

What's the correct way to go about computing the Inverse Laplace transform of this?

$$\frac{-2s + 1}{(s^2+2s+5)}$$

I Completed the square on the bottom but what do you do now?

$$\frac{-2s + 1}{(s+1)^2 + 4}$$

Answer 1

Hints:

the inverse Laplace Transforms of the two forms are:

$$\dfrac{a}{(s-b)^2 + a^2} = e^{bx} \sin ax$$

$$\dfrac{s-b}{(s-b)^2 + a^2} = e^{bx} \cos ax$$

Can you use those two forms and get a result of:

$$ e^{-t} \left(\dfrac{3}{2} \sin 2t - 2 \cos 2t\right)$$