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