<s>[INST] How find this integral $I=\int_{-\infty}^{+\infty}\frac{x^3\sin{x}}{x^4+x^2+1}dx$ Find this integral
$$I=\int_{-\infty}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
my idea: 
$$I=2\int_{0}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
because 
$$\dfrac{x^3\sin{x}}{x^4+x^2+1}\approx\dfrac{\sin{x}}{x},x\to\infty$$
so
$$I=\int_{-\infty}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
converges
then I can't,Thank you 
 [/INST] May I confess that I sawer nicer integrals ?  
Anyway, being patient and using Mhenni Benghorbal's suggestions plus a series of integrations by parts as well as a few changes of variables, the antiderivative is
$$\frac{1}{24} e^{-\sqrt[6]{-1}} \left(e^{\sqrt{3}} \left(\left(\sqrt{3}-3 i\right)
   \left(\text{Ei}\left(i x+(-1)^{5/6}\right)-\text{Ei}\left(\frac{1}{2} \left(-2 i
   x-\sqrt{3}+i\right)\right)\right)+\left(\sqrt{3}+3 i\right) e^i
   \left(\text{Ei}\left(\frac{1}{2} \left(-2 i
   x-\sqrt{3}-i\right)\right)-\text{Ei}\left(\frac{1}{2} \left(2 i
   x-\sqrt{3}-i\right)\right)\right)\right)-\left(\sqrt{3}-3 i\right) e^i
   \left(\text{Ei}\left(\frac{1}{2} \left(-2 i
   x+\sqrt{3}-i\right)\right)-\text{Ei}\left(\frac{1}{2} \left(2 i
   x+\sqrt{3}-i\right)\right)\right)+\left(\sqrt{3}+3 i\right)
   \left(\text{Ei}\left(\frac{1}{2} \left(-2 i
   x+\sqrt{3}+i\right)\right)-\text{Ei}\left(\frac{1}{2} \left(2 i
   x+\sqrt{3}+i\right)\right)\right)\right)$$ Once integrated between $-\infty$ and $\infty$, again after some patience, I arrived to $$\frac{\left(-i+\sqrt{3}+\left(\sqrt{3}+i\right) e^i\right) e^{-\sqrt[6]{-1}} \pi }{2
   \sqrt{3}}$$ that is to say $$-\frac{1}{3} e^{-\frac{\sqrt{3}}{2}} \pi  \left(\sqrt{3} \sin
   \left(\frac{1}{2}\right)-3 \cos \left(\frac{1}{2}\right)\right)$$ which reduces to $$\frac{2}{\sqrt{3}} e^{-\sqrt{3}/2} \pi \cos \frac{\pi+3}{6}$$ which corresponds to heropup's result.   
Thanks to Mhenni Benghorbal and heropup for their suggestions and ideas.
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