What is the inverse Laplace transform of (s+3)/(s^2+4s+4)?

Answer 1

Well, if I remember my Laplace transforms correctly,
L(f(x)) = ∫ [from t=0 to t=∞ of] f(x)e^(-st) dt

The above is first separated by partial fractions. Or better yet, write it as: (s+2+1)/(s+2)² = 1/(s+2) + 1/(s+2)²

Now, let's take the Laplace transform of f(t) = e^(-2t):
L(f(t)) = ∫ [from 0 to ∞ of] e^(-t(s+2)) dt
= -e^(-t(s+2))/(s+2) | 0 to infinity
At ∞ the exponential strongly goes to 0 so we are left with
= 1/(s+2)

Now, let's take the Laplace transform of f(t) = te^(-2t):
L(f(t)) = ∫ [from 0 to ∞ of] te^(-t(s+2)) dt
= -te^(-t(s+2))/(s+2) - e^(-t(s+2))/(s+2)² | from 0 to ∞
At ∞ the exponential strongly goes to 0, and at 0 the first term goes to 0, so we are left with:
1/(s+2)²

Thus, by inspection, the inverse Laplace transform of (s+3)/(s+2)² is
(t+1)e^(-2t) * u(t)
where that u(t) is the step function: 0 for negative t, 1 otherwise.

Answer 2

exp(-2t) + t*exp(-2t),

because (s + 3)/(s2 + 4s + 4) = 1/(s + 2) + 1/(s + 2)^2.