# Inverse Laplace Transform Assistance

How do I compute the following transform?
$$\frac {s-1}{2s^2+s+6}$$

I've gotten this far:

$$\frac {1}{2}\cdot \frac {s-1}{\left(s+\frac{1}{4}\right)^2 + \frac{47}{16}}$$

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You mean inverse transform, right? – Pedro Tamaroff Nov 21 '12 at 3:21
Yep sorry about that, I'll edit it. – Shankar Kumar Nov 21 '12 at 3:22

You can now use: $$\mathcal{L}_s\left(\mathrm{e}^{-\lambda t} \mathrm{e}^{i t \omega}\right) = \int_0^\infty \mathrm{e}^{-s t} \mathrm{e}^{-\lambda t} \mathrm{e}^{i t \omega} \mathrm{d} t = \frac{1}{s + \lambda - i \omega}$$ valid as long as $s+\lambda > 0$, and $\omega \in \mathbb{R}$. From here, taking real and imaginary parts you conclude: $$\mathcal{L}_s\left(\mathrm{e}^{-\lambda t} \cos\left(\omega t\right)\right) = \frac{s+\lambda}{(s+\lambda)^2 + \omega^2}, \qquad \mathcal{L}_s\left(\mathrm{e}^{-\lambda t} \sin\left(\omega t\right)\right) = \frac{\omega}{(s+\lambda)^2 + \omega^2}$$ Now you the decomposition you obtained and read off the coefficients, keeping in mind that the inverse Laplace transform has the form $\mathrm{e}^{-\lambda t} \left( \alpha \cos(\omega t) + \beta \sin(\omega t) \right)$ for some $\lambda, \omega, \alpha$ and $\beta$.
$$F(s)=\frac {s-1}{2s^2+s+6}=\frac {s-1}{2(s^2+\frac{s}{2} +3)}$$ $$ax^2+bx=a\left[ (x+\frac{b}{2a})^2-(\frac{b}{2a})^2 \right]$$ So: $$2(s^2+\frac{s}{2} +3)=2 \left[ (s+\frac{1}{4})^2-(\frac{1}{4})^2 +3 \right]=2 \left[ (s+\frac{1}{4})^2+(\frac{47}{15}) \right]$$ therefor: $$F(s)=(\frac{1}{2}) \frac{s-1 {\color{red} { +\frac{5}{4} -\frac{5}{4}} } }{(s+\frac{1}{4})^2+(\frac{47}{15})} = (\frac{1}{2}) \left[ \frac{s+\frac{1}{4}}{(s+\frac{1}{4})^2+(\frac{47}{15})}-\frac{\frac{5}{4}}{(s+\frac{1}{4})^2+(\frac{47}{15})} \right]$$ We know: $$\mathcal{L}\left[ e^{-at}cos(bt) \right]=\frac{s+a}{(s+a)^2+b^2}$$ $$\mathcal{L}\left[ e^{-at}sin(bt) \right]=\frac{b}{(s+a)^2+b^2}$$ therefor: $$f(t)=(\frac{1}{2})e^{\frac{-t}{4}} \left[ cos(\frac{\sqrt{47}}{4}t) - \frac{5}{\sqrt{47}} sin(\frac{\sqrt{47}}{4}t)\right]$$