So I'm trying to find the inverse laplace transform of $$\frac{1}{s^2+s+1}.$$
The first step I have taken is to change the equation to $$\frac{1}{(s+\frac{1}{2})^2+\frac{3}{4}}$$ by completing the square.
Then I use the formula: $e^{-at} = F(s+a)$ to get $$e^{-\frac{1}{2}t}\sin\left(\frac{\sqrt{3}}{2}\right).$$
This is where I am apparently taking the wrong step because the answer should be $$\frac{\sqrt{3}}{2}e^{-\frac{1}{2}t}\sin\left(\frac{\sqrt{3}}{2}\right).$$
So my question is, how do you get $\frac{\sqrt{3}}{2}$ in front of the equation?
$$F(s)=\frac{1}{s^2+s+1}.$$ $$F(s)=\frac{1}{(s+\frac 12)^2+\frac 34}.$$ $$F(s)=\dfrac {2}{\sqrt 3}\frac{\frac {\sqrt 3}{2}}{(s+\frac 12)^2+\frac 34}.$$ Apply inverse Laplace transform: $$f(t)=\dfrac {2}{\sqrt 3}e^{-t/2}\sin \left ( \dfrac {\sqrt 3 t}{2} \right)$$ Since $$ \mathcal {L^{-1}} \left (\dfrac {a}{s^2+a^2} \right) =\sin (at)$$