$$\int_{-\infty}^{\infty}{\frac{\sin x}{x^4-6x^2+10}\,\mathrm dx}$$
I get that when I evaluate the $\frac{\sin x}{x}$ one, I work with $\frac{e^{ix} - e^{-ix}}{2ix}$, I create a huge semicircle and a small one from $-\varepsilon$ to $\varepsilon$. Should I do the same here? And evaluate the integral of the small semicircle with $z=\varepsilon\cdot e^{i\varphi}$? And if the bigger semicircle gives zero, my integral equals to minus the small semicircle. Will I need the residues at all? Or is there a mistake somewhere?
Hint. Since $x^4-6x^2+10$ has no real zeros try to replace $\sin x$ with $\mathrm e^{\mathrm ix}$ via the Euler formula and take only the imaginary part of it like $$\int_{-\infty}^\infty\frac{\sin x}{x^4-6x^2+10}\,\mathrm dx = \operatorname{Im}\left(\int_{-\infty}^\infty\frac{\mathrm e^{\mathrm ix}}{x^4-6x^2+10}\,\mathrm dx\right)$$ since you will get a Fourier integral of the form $$\int_{-\infty}^\infty \frac{P(z)}{Q(z)}\mathrm e^{\mathrm iz}\,\mathrm dz=2\pi\mathrm i\sum_{z\in Q^{-1}(\{0\}),\operatorname{Im}z>0}\operatorname{Res}_{w=z}\left(\frac{P(w)}{Q(w)}\mathrm e^{\mathrm iw}\right).$$ However the given integral has value zero which can be shown in multiple ways.