V. Reshetnikov gave the remarkable integral, $$\int_0^1\frac{dx}{\sqrt[3]x\,\sqrt[6]{1-x}\,\sqrt{1-x\left(\sqrt{6}\sqrt{12+7\sqrt3}-3\sqrt3-6\right)^2}}=\frac\pi9(3+\sqrt2\sqrt[4]{27})\tag1$$ More generally, given some integer/rational $N$, we are to find an algebraic number $\alpha$ that solves,
$$\int_0^1\frac{dx}{\sqrt[3]x\ \sqrt[6]{1-x}\ \sqrt{1-x\,\alpha^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{3}\,|\alpha|}\tag2$$
and absolute value $|\alpha|$. (Compare to the similar integral in this post.) Equivalently, to find $\alpha$ such that,
$$\begin{aligned} \frac{1}{N} &=I\left(\alpha^2;\ \tfrac12,\tfrac13\right)\\[1.8mm] &= \frac{B\left(\alpha^2;\ \tfrac12,\tfrac13\right)}{B\left(\tfrac12,\tfrac13\right)}\\ &=B\left(\alpha^2;\ \tfrac12,\tfrac13\right)\frac{\Gamma\left(\frac56\right)}{\sqrt{\pi}\,\Gamma\left(\frac13\right)}\end{aligned} \tag3$$
with beta function $\beta(a,b)$, incomplete beta $\beta(z;a,b)$ and regularized beta $I(z;a,b)$. Solutions $\alpha$ for $N=2,3,4,5,7$ are known. Let, $$\alpha=\frac{-3^{1/2}+v^{1/2}}{3^{-1/2}+v^{1/2}}\tag4$$ Then, $$ - 3 + 6 v + v^2 = 0, \quad N = 2\\ - 3 + 27 v - 33v^2 + v^3 = 0, \quad N = 3\\ 3^2 - 150 v^2 + 120 v^3 + 5 v^4 = 0, \quad N = 5\\ - 3^3 - 54 v + 1719 v^2 - 3492v^3 - 957 v^4 + 186 v^5 + v^6 = 0, \quad N = 7$$
and (added later),
$$3^4 - 648 v + 1836 v^2 + 1512 v^3 - 13770 v^4 + 12168 v^5 - 7476 v^6 + 408 v^7 + v^8 = 0,\quad N=4$$
using the largest positive root, respectively. The example was just $N=2$, while $N=4$ leads to,
$$I\left(\tfrac{1-\alpha}{2};\tfrac{1}{3},\tfrac{1}{3}\right)=\tfrac{3}{8},\quad\quad I\left(\tfrac{1+\alpha}{2};\tfrac{1}{3},\tfrac{1}{3}\right)=\tfrac{5}{8}$$
I found these using Mathematica's FindRoot command, and some hints from Reshetnikov's and other's works, but as much as I tried, I couldn't find prime $N=11$.
Q: Is it true one can find algebraic number $\alpha$ for all $N$? What is it for $N=11$?
(Too long for a comment. And courtesy of V. Reshetnikov's result here, though as he points out it is tentative.)
The algebraic number $\alpha$ that solves,
$$\int_0^1\frac{dx}{\sqrt[3]x\ \sqrt[6]{1-x}\ \sqrt{1-x\,\alpha^2}}=\frac{1}{11}\,\frac{2\pi}{\sqrt{3}\;\alpha}$$
seems to have a $40$-deg minpoly. However, it turns out we can also reduce its degree and express it using the common form above. Let,
$$\alpha=\frac{3^{1/2}-v^{1/2}}{3^{-1/2}+v^{1/2}}$$
where $v$ is a the second largest positive root ($r_9$ in Mathematica syntax) of,
$$\small P(v)=-3^{10} + 23816430 v^2 - 323903448 v^3 + 2177615583 v^4 - 9297934272 v^5 + 25869358152 v^6 - 37475802144 v^7 - 16459141842 v^8 + 180065426112 v^9 - 338100745356 v^{10} + 329418595440 v^{11} - 211367836746 v^{12} + 102243404736 v^{13} - 8162926200 v^{14} - 9999738144 v^{15} + 1006439643 v^{16} - 134177472 v^{17} - 2246706 v^{18} + 30888 v^{19} + 11 v^{20} = 0$$
Also,
$$I\big(\alpha^2;\tfrac{1}{2},\tfrac{1}{3}\big)=\tfrac{1}{5}\,I\big(\tfrac{1-\alpha}{2};\tfrac{1}{3},\tfrac{1}{3}\big)=\tfrac{1}{6}\,I\big(\tfrac{1+\alpha}{2};\tfrac{1}{3},\tfrac{1}{3}\big)=\frac{1}{11}$$
with regularized beta function $I(z;a,b)$. Furthermore, if
$$y =\frac{r_1+r_9+r_{13}+r_{14}}{12}$$
then $y$ is a root of the solvable quintic,
$$67 - 1748 y - 7033 y^2 - 1378 y^3 + 234 y^4 + y^5=0$$
with discriminant divisible by $11^4\times23^4$. Using the other quartic symmetric polynomials show that the $20$-deg is just a quartic in disguise, hence is solvable. All these suggest that $P(v)$ is the correct polynomial for $N=11$.