Find the following reverse Laplace: $$L^{-1}\left\{ \dfrac{s}{s^4-s^2+1} \right\}(t)$$ I search in internet and get a result $\dfrac{2\sin\left(\tfrac{t}{2}\right)\sinh \left(\tfrac{\sqrt{3}t}{2} \right)}{\sqrt{3}}$. I tried some ways, but it's useless. I don't understand why product of $\sin$ and $\sinh$ appear here.
I hope to have hints from you. Thank you.
Following your transformation, you have \begin{align*} L^{-1}\left[\frac{s}{s^4-s^2+1}\right] &=L^{-1}\left[\frac{s}{((s-\sqrt{3}/2)^2+1/4)((s+\sqrt{3}/2)^2+1/4)}\right]\\ &=\frac{1}{2\sqrt{3}}\,L^{-1}\left[\frac{1}{((s-\sqrt{3}/2)^2+1/4)}-\frac{1}{((s+\sqrt{3}/2)^2+1/4)}\right]\\ &=\frac{1}{2\sqrt{3}}\left[2e^{\sqrt{3}\,t/2}\sin(t/2)-2e^{-\sqrt{3}\,t/2}\sin(t/2)\right]\\ &=\frac{1}{\sqrt{3}}\,\sin(t/2)\,\left[e^{\sqrt{3}\,t/2}-e^{-\sqrt{3}\,t/2}\right]\\ &=\frac{2}{\sqrt{3}}\,\sin(t/2)\,\left[\frac{e^{\sqrt{3}\,t/2}-e^{-\sqrt{3}\,t/2}}{2}\right]\\ &=\frac{2}{\sqrt{3}}\,\sin(t/2)\,\sinh(\sqrt{3}\,t/2), \end{align*} as required.