Originally Posted by

**Mush** Err, I think you've got your completing the square wrong.

$\displaystyle s^2 + 6s + 9 \neq (s+3)^2 + 4 $

$\displaystyle (s+3)^2 + 4 = s^2 + 6s + 9 + 4 = s^2 + 6s + 13.$

The way I see it is:

$\displaystyle \frac{s+9}{s^2 + 6s + 9} = \frac{s+9}{(s+3)^2} = \frac{s+3}{(s+3)^2} + \frac{6}{(s+3)^2} = \frac{1}{s+3} + 6 \bigg(\frac{1}{(s+3)^2} \bigg) $

Then for the inverse, remember the following rules:

$\displaystyle \mathcal{L}^{-1} \bigg( \frac{1}{s+a}\bigg) = e^{-at} $

$\displaystyle \mathcal{L}^{-1} \bigg(\frac{n!}{(s+a)^{n+1}} \bigg)= t^n e^{-at} $