Inverse Laplace transform of $\frac{s^2-4s-2}{\left(s^2+2\right)^2}$

Question

I approached this problem as follow:

$1.$ rewrote $(s^2-4s-2)$ into $(s-2)^2-6$

$2.$ Now break the function into 2 parts: $\frac{(s-2)^2}{(s^2+2)^2} + \frac{6}{(s^2+2)^2}$

the Laplace inverse for $\frac{w}{s^2+w^2}$ is $\sin(w*t)$

and the Laplace inverse for $\frac{s}{s^2+w^2}$ is $\cos(w*t)$

but i can't get rid of those powers to the 2 in the question.

please help.

Answer 1

Hint. You may use partial fraction $$ \frac{s^2-4s-2}{\left(s^2+2\right)^2}=\frac{1}{2+s^2}-\frac{4 s+4}{\left(2+s^2\right)^2} $$ and then make use of a table.