I. Integrals
We have (typo corrected),
\begin{align} I_1 &=\pi =\int_{-\infty}^{\infty}\frac{(x-1)^2}{\color{blue}{(2x - 1)}^2 + (x^2 - x)^2}\,dx,\quad\text{(by Mark S.)}\\[1.8mm] I_3 &=\pi =\int_{-\infty}^{\infty}\frac{(x+1)^2}{\color{blue}{(x + 1)}^2 + (x^2 + x)^2}\,dx\\[1.8mm] I_5 &=\pi =\int_{-\infty}^{\infty}\frac{(x+1)^2}{\color{blue}{(x^2 - x - 1) }^2 + (x^2 + x)^2}\,dx\\[1.8mm] \color{red}{I_7} &=\pi =\int_{-\infty}^{\infty}\frac{(x+1)^2}{\color{blue}{(x^3 + 2x^2 - x - 1)}^2 + (x^2 + x)^2}\,dx\\[1.8mm] I_9 &=\pi =\int_{-\infty}^{\infty}\frac{(x-1)^2}{\color{blue}{(x^3 - 3x^2 + 1)}^2 + (x^2 - x)^2}\,dx\\[1.8mm] I_{11} &=\, ?? =\int_{-\infty}^{\infty}\frac{(x\pm1)^2}{\color{blue}{(x^5 + 3x^4 - 3x^3 - 4x^2 + x + 1)}^2 + (x^2 \pm x)^2}\,dx \end{align}
where those in blue are the minimal polynomials of $x=\frac{1}{2\cos(2\pi/p)}$ for $p=1,3,5,7,9,11$. These integrals have the form,
$$I_p =\int_{-\infty}^{\infty}\frac{(x\pm 1)^2}{F_p(x)}$$
where $F(x)=0$ is an equation with a solvable Galois group excepting $p=11$. The red integral $I_7$ is the one in the post, A nasty integral of a rational function,
$$\int_0^{\infty} \frac{x^8 - 4x^6 + 9x^4 - 5x^2 + 1}{x^{12} - 10 x^{10} + 37x^8 - 42x^6 + 26x^4 - 8x^2 + 1} \, dx = \frac{\pi}{2}$$ as well as in this post after some manipulation.
II. Question 1
Q: Why did the "pattern" of using minimal polynomials work then stop at $p=11$, and how can we make it continue by adjusting other parameters?
III. Alternative forms
Based on an insight from an old post, for $p=7$ we use the "negative" case on both its numerator and denominator. The denominator is a sextic again with a solvable Galois group, discriminant factor $12833,$ and we find,
$$\int_{-\infty}^{\infty}\frac{(x\color{red}-1)^2}{\color{blue}{(x^3 + 2x^2 - x - 1)}^2 + (x^2 \color{red}- x)^2}\,dx=\pi\sqrt{\frac{u}{\color{green}{12833}}}$$
where $u$ is a root of a nonic also with a solvable Galois group,
$$\small -\color{green}{12833}^3*1782434241^2 - 41120374319577904376201744753 u - 354521093943488815427187669 u^2 - 550802363395052799639795 u^3 - 176617825075778391189 u^4 + 116970252692553921 u^5 - 20201478347596 u^6 + 1625465206 u^7 - 63997 u^8 + u^9=0$$
For $p=9$, if we use the positive case, the denominator still is solvable. However, for $p=11$, then $F(x) = 0$ is not solvable for either case.
IV. Question 2
Q: So was the pattern interrupted because the denominator of $p=11$ no longer has a solvable Galois group?
