Question: Find inverse Laplace transform of: [9/(((s^2)+9)^2)]; giving every step.
Using a table of (inverse) Laplace Transforms,
f(t) = sin(at) - at.cos(at) gives L{f(t)} = F(s) = 2a^3/(s^ + a^2)^2
setting a = 3,
f(t) = sin(3t) - 3t.cos(3t) gives L{f(t)} = F(s) = 2.3^3/(s^ + 3^2)^2 = 2*27/(s^2 + 9)^2
f(t) = sin(3t) - 3t.cos(3t) gives L{f(t)} = 6*9/(s^2 + 9)^2
Hence,
f(t) = (1/6)(sin(3t) - 3t.cos(3t)) gives L{f(t)} = 9/(s^2 + 9)^2
Answer: The inverse Laplace Transform of 9/(s^2 + 9)^2 is: (1/6)(sin(3t) - 3t.cos(3t))