Find the inverse Laplace of the function F(s)= (s^2 − 5)/ (s^3 + s^2 + 9s + 9)

F(s)= (s^2 − 5)/ (s^3 + s^2 + 9s + 9)

I'm pretty much lost here...will appreciate any help :))

The denominator can be factored by grouping:
[s³ + 9s] + [s² + 9]
s[s² + 9] + [s² + 9]
(s + 1)(s² + 9)

Let's use partial fractions to split this up:
s² - 5 = A(s² + 9) + B(s + 1)
If s = -1, A = -2/5
Simplifying:
(7/5)s² - 7/5 = B(s + 1)
(7/5)(s + 1)(s - 1) = B(s + 1)
Therefore B = (7/5)(s - 1)

F(s) = (7/5)∙(s - 1)/(s² + 9) - (2/5)∙1/(s + 1)

You should know that
ℒ{exp(at)∙f(t)} = F(s - a)
ℒ{sin(at)} = a/(s² + a²)
ℒ{cos(at)} = s/(s² + a²)
ℒ{1} = 1/s

Therefore, we can split the first term into sine and cosine, and the second term into a frequency shift.

(7/5)∙ℒ⁻¹{s/(s² + 9) - (1/3)*3/(s² + 9)} - (2/5)∙e^(-t)∙ℒ⁻¹{1/s}

(7/5)∙[cos(3s) - (1/3)sin(3s)] - (2/5)∙e^(-t)