Inverse Laplace transform - using the table I am trying to find the inverse Laplace transform $(g(t))$ of
$$ G(s) =  \frac{2s}{(s+1)^2+4}$$
I know about the inverse transforms $e^{a t}\cos(\omega t)$ and $\mathrm{e}^{at}\sin(\omega t)$ however I am trying to get the inverse transforms without these, as these are not on the table of standard transforms for our course.
I was also attempting to use $\mathrm{e}^{at} f(t) \leftrightarrow F(s-a)$, but I'm not sure how to go about this, either.
Any help would be appreciated. :)

If you have $(s+1)^2$ in the denominator then make it appear in the numerator by adding zero:
$$ G(s) = \frac{2s}{(s+1)^2+4} = \frac{2(s+1-1)}{(s+1)^2+4} = 2 \frac{s+1}{(s+1)^2+4}-\frac{2}{(s+1)^2+4} $$
