How would I solve this inverse laplace transform?

Question

(s+1)/(4s^2 + 9)

I just can't get my head around what to do about the 4s^2.

Answer 1

F(s) = (s+1)/(4s^2 + 9) =
Extract the 4 from 4s^2:
(1/4)(s+1)/(s^2 + (3/2)^2) =
Separate the fractions:
(1/4)(s/(s^2 + (3/2)^2) + 1/(s^2 + (3/2)^2) =
Change the form of the 2nd term to a standard form:
(1/4)[s/(s^2 + (3/2)^2) + (2/3)(3/2)/(s^2 + (3/2)^2)]
Look up (or integrate) the transforms:

f(t) = (1/4)u(t)[cos((3/2)t) + (2/3)cos((3/2)t)]

Change to function w/ phase angle form:
tan(φ) = 2/3
f(t) = ((√13)/12)u(t)[cos((3/2)t) - arctan(2/3)]

Answer 2

answer is 4s^3 + 9s + 4s² + 9

Answer 3

= 1/4( s/(s^2+9/4)+1/s^2+9/4)
1/4[cos(3/2*t)+2/3*sin(3/2*t)]