Inverse Laplace with a irreducible quadratic in the denominator and a 1 in the numerator

Question

Please help me find the Inverse Laplace transform of: $$F(s) = \frac{1}{s(s^2+8s+4)}$$ After completing the square, I obtained $$F(s) = \frac{1}{s((s+4)^2-12)}$$

Thank you.

Answer 1

Using partial fraction decomposition gives $$F(s) = \frac{1}{s(s^2+8s+4)}=\frac{1}{4}\bigg(\frac{1}{s}-\frac{s}{s^2+8s+4}-\frac{8}{s^2+8s+4}\bigg)$$ $$\therefore F(s)=\frac{1}{4}\bigg(\frac{1}{s}-\frac{s}{(s+4)^2-12}-\frac{8}{(s+4)^2-12}\bigg)$$ Then in order to use $$\mathcal{L}\{e^{at}\}=\frac{1}{s-a}$$ $$\mathcal{L}\{e^{at}\sinh{(bt)}\}=\frac{b}{(s-a)^2-b^2}$$ $$\mathcal{L}\{e^{at}\cosh{(bt)}\}=\frac{s-a}{(s-a)^2-b^2}$$ we can rewrite $F(s)$ as $$F(s)=\frac{1}{4}\bigg(\frac{1}{s}-\frac{s+4}{(s+4)^2-12}-\frac{4}{(s+4)^2-12}\bigg)$$ $$=\mathcal{L}\bigg\{\frac{1}{4}\bigg(1-e^{-4t}\cosh{(\sqrt{12}t)}-\frac{4}{\sqrt{12}}e^{-4t}\sinh{(\sqrt{12}t)}\bigg)\bigg\}$$