Find the inverse laplace transform of $F(s) = \frac{2s}{s^{4}+s^{2}+1}$

Question

I need to find the inverse Laplace transform of $F(s) = \frac{2s}{s^{4}+s^{2}+1}.$

I realise that I need to do something with the denominator so I can convert to partial fractions, but I am not able to do so. Any help will be appreciated, thanks!

Answer 1

HINT:

$$s^4+s^2+1=(s^2+1)^2-s^2=\cdots$$

$$\implies2s=s^2+s+1-(s^2-s+1)$$

Answer 2

Since: $$\mathcal{L}^{-1}\left(\frac{s}{s^2-a^2}\right)=\cosh(at),\qquad\mathcal{L}^{-1}\left(\frac{a}{s^2+a^2}\right)=\sin(at)\tag{1}$$ and: $$\frac{2s}{s^4+s^2+1} = \frac{1}{s^2-s+1}-\frac{1}{s^2+s+1}\tag{2}$$ it follows that: $$\mathcal{L}^{-1}(F(s)) = \color{red}{\frac{4}{\sqrt{3}}\cdot\sinh\left(\frac{t}{2}\right)\cdot\sin\left(\frac{\sqrt{3}}{2}t\right)}.\tag{3}$$