Question

inverse laplace transform F(s)=2s+12/s^2+2s+5

Answer 1

\[F(s)=\frac{2s+12}{s^2+2s+5}\]
I would first complete the square in the denominator, then split up into two fractions:
\[s^2+2s+5\\
s^2+2s+1+4\\
(s+1)^2+4\]
\[F(s)=\frac{2s}{(s+1)^2+4}+\frac{12}{(s+1)^2+4}\]
If you have a table of inverse transforms near you, it helps to rewrite the above as
\[F(s)=\frac{2s+2-2}{(s+1)^2+4}+\frac{12}{(s+1)^2+4}\\
F(s)=2\left(\frac{s+1}{(s+1)^2+4}\right)+\frac{10}{(s+1)^2+4}\\
F(s)=2\left(\frac{s+1}{(s+1)^2+4}\right)+5\left(\frac{2}{(s+1)^2+4}\right)\]
Does that help?

Answer 2

@raprap, if you don't have a table of inverse transforms, you should know that, if \(\mathcal{L}\{f(t)\}=F(s)\), then
\[\mathcal{L}^{-1}\left\{F(s-a)\right\}=e^{at}f(t)\\
\mathcal{L}^{-1}\left\{\frac{s}{s^2+b^2}\right\}=\cos t\\
\mathcal{L}^{-1}\left\{\frac{b}{s^2+b^2}\right\}=\sin t\]