An interesting integral

This is a problem related to A problem in complex analysis. $$\int_{-\infty}^\infty \dfrac{dx}{x^4+x^3+x^2+x+1}=\dfrac{4\pi}{5}\sin \dfrac{2\pi}{5}$$

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Method I: Real analysis
Letting $a=\cos \dfrac{2\pi}{5},b=\cos \dfrac{4\pi}{5}$, we have $$\begin{align}\int_{-\infty}^\infty \dfrac{dx}{x^4+x^3+x^2+x+1}&=\int_{-\infty}^\infty \dfrac{1}{(x^2-2x\cos \dfrac{2\pi}{5}+1)(x^2-2x\cos \dfrac{4\pi}{5}+1)}dx\\&=\dfrac{1}{2(a-b)}\int_{-\infty}^\infty \bigg(\dfrac{-x+2a}{x^2+1-2xa}-\dfrac{-x+2b}{x^2+1-2xb}\bigg)dx\\&=\dfrac{1}{2(a-b)}\bigg(\dfrac{1}{2}\ln \bigg|\dfrac{x^2+1-2xb}{x^2+1-2xa}\bigg| \bigg]_{-\infty}^\infty\\& +\int_{-\infty}^\infty \dfrac{a}{x^2+1-2xa}dx-\int_{-\infty}^\infty \dfrac{b}{x^2+1-2xb}dx\bigg)\\&=\dfrac{1}{2(a-b)}\bigg(0+\int_{-\infty}^\infty \dfrac{a}{x^2+1-2xa}dx-\int_{-\infty}^\infty \dfrac{b}{x^2+1-2xb}dx\bigg)\\&=\dfrac{1}{2(a-b)}\bigg(\int_{-\infty}^\infty \dfrac{a}{(x-a)^2+1-a^2}dx-\int_{-\infty}^\infty \dfrac{b}{(x-b)^2+1-b^2}dx\bigg)\\&=\dfrac{\pi}{2(a-b)} \bigg( \dfrac{a}{\sqrt{1-a^2}}-\dfrac{b}{\sqrt{1-b^2}}\bigg)\\&=\dfrac{\pi}{\sqrt 5}\bigg(\dfrac{\cos \dfrac{2\pi}{5}}{\sin \dfrac{2\pi}{5}}-\dfrac{\cos \dfrac{4\pi}{5}}{\sin \dfrac{4\pi}{5}}\bigg)\\&=\dfrac{\pi}{\sqrt 5} \dfrac{\sin \dfrac{2\pi}{5}}{\sin \dfrac{2\pi}{5}\sin \dfrac{4\pi}{5}}\end{align}$$ It remains to prove that $\sqrt{5}\sin \dfrac{2\pi}{5}\sin \dfrac{4\pi}{5}=\dfrac{5}{4}$. We have $$\begin{align}\sqrt{5}\sin \dfrac{2\pi}{5}\sin \dfrac{4\pi}{5}&=2\sqrt{5}\sin^2 \dfrac{2\pi}{5} \cos \dfrac{2\pi}{5}\\&=2\sqrt{5}(\cos \dfrac{2\pi}{5}-\cos^3 \dfrac{2\pi}{5})\\&=2\sqrt{5}\dfrac{\sqrt{5}-1}{4}\bigg(1-(\dfrac{\sqrt{5}-1}{4})^2\bigg)\\&=\dfrac{\sqrt{5}(\sqrt{5}-1)(10+2\sqrt{5})}{32}\\&=\dfrac{5}{4}.\end{align}$$ Finally, $$\int_{-\infty}^\infty \dfrac{dx}{x^4+x^3+x^2+x+1}=\dfrac{4\pi}{5}\sin \dfrac{2\pi}{5}.$$ Sorry for the clumsy solution. I venture there should be a much faster way... But it's nice to review so many techniques in one problem!
Remark: We have used nth roots of unity, partial fraction decomposition (algebraic identity), concept of limit, tangent substitution, trigonometry identity, value of $\cos \dfrac{2\pi}{5}$ and $\cos \dfrac{4\pi}{5}$.

Method II: Complex analysis residue theorem
$$\int_{-\infty}^\infty \dfrac{P(z)}{Q(z)}\;dz=2\pi i\sum_{i=1}^m \dfrac{P(a_i)}{Q'(a_i)},$$ where $a_1,\cdots,a_m$ are the poles of $P/Q$ in the upper half plane. $$\begin{align}\int_{-\infty}^\infty \dfrac{dx}{x^4+x^3+x^2+x+1}&=\int_{-\infty}^\infty \dfrac{x-1}{x^5-1}\; dx\\&=2\pi i\bigg(\dfrac{a-1}{5a^4}+\dfrac{a^2-1}{5a^8}\bigg)\\&=\dfrac{2\pi i}{5}(a^2-a+a^{-1}-a^2)\\&=\dfrac{2\pi i}{5}(-2i\sin \dfrac{2\pi}{5})\\&=\dfrac{4\pi}{5}\sin \dfrac{2\pi}{5},\end{align}$$ where $a=e^{2\pi i/5}$ and $a^5=1$.