find the inverse laplace transform of: $f(s)=\\frac{1}{s^2 + 4s + 5}$

Explanation:

Step1: Complete the square in denominator

$s^2 + 4s + 5 = (s+2)^2 + 1$ So $F(s) = \frac{1}{(s+2)^2 + 1}$

Step2: Apply inverse Laplace shift theorem

Recall $\mathcal{L}^{-1}\left{\frac{1}{s^2 + a^2}\right} = \sin(at)$, and $\mathcal{L}^{-1}{F(s-c)} = e^{ct}f(t)$ Here $c=-2$, $a=1$, $f(t)=\sin(t)$

Answer:

$e^{-2t}\sin(t)$