Can This Complex Integral Be Proven Rational?

[anemone]

Here is this week's POTW:

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Prove that \(\displaystyle \int_{-100}^{-10} \left(\frac{x^2-2}{x^3-3x+1}\right)^2\,dx+\int_{\frac{1}{101}}^{\frac{1}{11}} \left(\frac{x^2-2}{x^3-3x+1}\right)^2\,dx + \int_{\frac{101}{100}}^{\frac{11}{10}} \left(\frac{x^2-2}{x^3-3x+1}\right)^2\,dx \) is a rational number.

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[anemone]

No one answered last week POTW.:( You can find the suggested solution as follows:

Spoiler

The polynomial $x^3-3x+1$ changes sign in each of the interval $\left[-2,\,-1\right],\,\left[\dfrac{1}{3},\,\dfrac{1}{2}\right],\,\left[\dfrac{3}{2},\,2\right]$. So it has no zeroes outside these intervals. Hence the integral in question is continuous on the three ranges of integration.

By the substitution $x=\dfrac{1}{1-t}$ and $x=1-\dfrac{1}{t}$, the intervals over $\left[\dfrac{1}{101},\,\dfrac{1}{11}\right]$ and $\left[\dfrac{101}{100},\,\dfrac{11}{10}\right]$ are respectively converted into integrals over $\left[-100,\,-10\right]$.

The integrand

$P(x)=\left(\dfrac{x^2-x}{x^3-3x+1}\right)^2$

is invariant under each of the substitution $x\rightarrow \dfrac{1}{1-x}$ and $x\rightarrow 1-\dfrac{1}{x}$.

Hence the sum of the three given integrals is expressible as

\(\displaystyle \int_{-100}^{-10} \left(\frac{x^2-2}{x^3-3x+1}\right)^2 \left(1+\dfrac{1}{x^2}+\dfrac{1}{(1-x)^2}\right)\,dx \)

But

$\dfrac{1}{P(x)}=\left(x+1-\dfrac{1}{x}-\dfrac{1}{x-1}\right)^2$

So the last integral is of the form \(\displaystyle \int u^{-2}\,du \).

Hence, its value is

$ \left.-\dfrac{x^2-x}{x^3-3x+1}\right|_{-100}^{-10}$, which is rational.