Inverse Laplace Transform Assistance How do I compute the following transform?
$$\frac {s-1}{2s^2+s+6}$$
I've gotten this far: 
$$\frac {1}{2}\cdot \frac {s-1}{\left(s+\frac{1}{4}\right)^2 + \frac{47}{16}}$$
 A: 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 $\beta$.
A: $$ F(s)=\frac {s-1}{2s^2+s+6}=\frac {s-1}{2(s^2+\frac{s}{2} +3)}  $$
$$ ax^2+bx=a\left[  (x+\frac{b}{2a})^2-(\frac{b}{2a})^2 \right] $$
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:
$$ f(t)=(\frac{1}{2})e^{\frac{-t}{4}} \left[ cos(\frac{\sqrt{47}}{4}t)  - \frac{5}{\sqrt{47}} sin(\frac{\sqrt{47}}{4}t)\right]   $$
