# Use Laplace transforms to find $f(t)$

this problem has been giving me a bit of trouble and was posting it to here to help me get some clarification. I know that I have to complete the square, but I end up getting lost at that point. Any help is much appreciated!

$$L^{-1} \big\{\frac{s}{s^2+6s+10}\big\}$$

$$F(s) = \frac{s}{s^2+6s+10}$$
$$\frac{s+3-3}{(s+3)^2+1} = \frac{s+3}{(s+3)^2+1} - \frac{3}{(s+3)^2+1}$$
$$f(t) = L^{-1} \{F(s)\} = e^{-3t} \cos t - 3e^{-3t} \sin t$$