On Reshetnikov's integral $\int_0^1\frac{dx}{\sqrt[3]x\ \sqrt[6]{1-x}\ \sqrt{1-x\,\alpha^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{3}\,|\alpha|}$ 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 $N=11$? 

 A: 
I. Duplication

Following Nemo's lead in this answer, we find the formula,
$$\frac{1}{2}I(p^2;\tfrac{1}{2},\tfrac{1}{3})=I(1+q^3;\tfrac{1}{2},\tfrac{1}{3})$$ 
where $p,q$ are related by the $12$-deg,
$$p^2(-2 + 2 q + q^2)^6 = 36(1 + q^3) (4 + 4 q + 6 q^2 - 2 q^3 + q^4)^2$$
This then enables us to find infinitely many $\displaystyle\frac{1}{2^n N}$.
For example, since $I(p^2;\tfrac{1}{2},\tfrac{1}{3})=\frac{1}{3}$ is known, then solving for $I(\alpha^2;\tfrac{1}{2},\tfrac{1}{3})=\frac{1}{6}$ turns out to involve a $36$-deg equation.

II. Triplication

(Courtesy of Nemo.) Starting with,
$$B\left(z;\frac{1}{2},\frac{1}{3}\right)=2 \sqrt{z} \, _2F_1\left(\frac{1}{2},\frac{2}{3};\frac{3}{2};z\right).
$$
The transformation
$$
\, _2F_1\left(\frac{1}{2},\frac{2}{3};\frac{3}{2};-\frac{3 z \left(1-\frac{z}{9}\right)^2}{(1-z)^2}\right)=\frac{(1-z) \, }{1-\frac{z}{9}}{}_2F_1\left(\frac{1}{2},\frac{2}{3};\frac{3}{2};z\right)
$$
applied two times gives
$$
\frac{1}{3} B\left({\frac{(9-z)^2 z \left(z^3+225 z^2-405 z+243\right)^2}{729 (1-z)^2 (z+3)^6}};\frac{1}{2},\frac{1}{3}\right)=B\left(z;\frac{1}{2},\frac{1}{3}\right).
$$
