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Is the information below correct?

Find the inverse Laplace transform of $$ F(s) = \frac{s}{s^2 + 4s + 13}$$

Soln: a) Complete the squares to simplify our denominator $$ s^2 + 4s + 13 = (s+2)^2 + 9 = (s+2)^2 + 3^2$$ $$\mathscr{L}^{-1}\left\{F(s)\right\} = \frac{s}{(s+2)^2 + 3^2}. $$ From the table we can deduce that this is $$\mathscr{L}^{-1}\left\{F(s)\right\} = e^{-2t} \cos(3t).$$

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    $\begingroup$ (+1) for showing your work, among other things. Don't forget to "accept" an answer when it's all said and done! You can also revisit old questions of yours and accept answers, if you're into the whole "closure" thing :) $\endgroup$ – The Chaz 2.0 Jul 20 '11 at 4:17
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HINT: $$ \frac{s}{{(s + 2)^2 + 3^2 }} = \frac{{s + 2 - 2}}{{(s + 2)^2 + 3^2 }} = \frac{{s + 2}}{{(s + 2)^2 + 3^2 }} - \frac{2}{3}\frac{3}{{(s + 2)^2 + 3^2 }}. $$ Now note that the inverse transforms of $\frac{{s + \alpha }}{{(s + \alpha )^2 + \omega ^2 }}$ and $\frac{\omega}{{(s + \alpha)^2 + \omega^2 }}$ are $e^{-\alpha t} \cos(\omega t)$ and $e^{-\alpha t} \sin (\omega t)$, respectively.

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    $\begingroup$ Shoot!!! I see what you mean. Thanks ! $\endgroup$ – user10695 Jul 20 '11 at 3:04
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For rational functions you can use decomposition to partial fractions: $$\frac{s}{(s+2)^2+3^2}=\frac{s}{(s+2+3i)(s+2-3i)}=\frac{1/2+i/3}{s+2-3i}+\frac{1/2-i/3}{s+2+3i}.$$ The only thing you need to know now is that the inverse Laplace transform of $1/(s+\alpha)$ is $\exp(-\alpha x)$.

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