I have noticed experimentally that: $$\int_0^1 \frac{\color{red}{x}}{x^4+2x^3+2x^2-2x+1} \,dx=\color{blue}{\frac{\pi}{8}},\tag{1}$$ $$\int_0^1 \frac{\color{red}{1-x^2}}{x^4+2x^3+2x^2-2x+1} \, dx=\color{blue}{\frac{\pi}{4}},\tag{2}$$ $$\int_0^1 \frac{\color{red}{1+x-x^2}}{x^4+2x^3+2x^2-2x+1} \, dx =\color{blue}{\frac{3\pi}{8}}.\tag{3}$$ So slight variations in the numerator always seem to produce something like $n\pi$, where $n$ is a rational number. At MathSE I have asked what the exact relationship is between $n$ and the numerator of the integrand, to which Quanto has responded with a general formula: $$\int_{0}^{1} \frac{ax^2 +b x + c}{x^4+2x^3+2x^2-2x+1} \, dx =\frac\pi8(c+b-a)+\frac\pi{3\sqrt3}(a+c).\tag{4}$$ However, is it necessary for the denominator to remain fixed? Not really. The following integral is formula (34) in this list of $\pi$ formulas: $$\int_0^1 \frac{\color{red}{16x-16}}{x^4-2x^3+4x-4}\,dx=\color{blue}{\pi}.\tag{5}$$ Notice that the denominator is different. But again, a slight variation in the denominator and it still produces something like $n\pi$: $$\int_0^1 \frac{\color{red}{x^2-x-1}}{x^4-2x^3+4x-4}\,dx=\color{blue}{\frac{3\pi}{16}}.\tag{6}$$ The fact that the denominator is not the same suggests that a further generalization is possible.

Here is my question: is it possible to characterize the integrand $\frac{P(x)}{Q(x)}$ in such a way that by simple inspection we can say $I=n\pi$? Or in other words, what should be the relationship between the coefficients of the numerator and the denominator for the integral to yield $n\pi$?