On the integral $\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt{1-x}\ \sqrt[4]{1-x\,\gamma^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{2\gamma}}$ V. Reshetnikov gave the interesting integral,
$$\int_0^1\frac{\mathrm dx}{\sqrt[4]x\ \sqrt{1-x}\ \sqrt[4]{2-x\,\sqrt3}}=\frac{2\,\sqrt2}{3\,\sqrt[8]3}\pi\tag1$$
After some experimentation, it turns out that more generally, given some integer/rational $N$, we are to find an algebraic number $\gamma$ that solves,

$$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt{1-x}\ \sqrt[4]{1-x\,\gamma^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{2\gamma}}\tag2$$

(Compare to the similar integral in this post.) Equivalently, to find $\gamma$ such that,
$$\begin{aligned}
\frac{1}{N}
&=I\left(\gamma^2;\ \tfrac14,\tfrac14\right)\\[1.8mm]
&= \frac{B\left(\gamma^2;\ \tfrac14,\tfrac14\right)}{B\left(\tfrac14,\tfrac14\right)}
\end{aligned} \tag3$$
with beta function $B(a,b)$, incomplete beta $B(z;a,b)$ and regularized beta $I(z;a,b)$, and $B\left(\tfrac14,\tfrac14\right)=\frac{\sqrt\pi}{\Gamma^2\left(\frac14\right)}$. Reshetnikov's example, after tweaking, was just the case $N=\frac{3}{2}$ and $\gamma=\frac{3^{1/4}}{\sqrt{2}}$. 
Solutions for prime $N=2,3,5,7$ are known. Let $v=\gamma$, then,
$$-1 + 2 v^2 = 0\quad\quad  N=2\\ - 1 + 2 v + 2 v^2 = 0\quad\quad  N=3\\ - 1 + 8 v - 4 v^2 - 8 v^3 + 4 v^4 = 0\quad\quad  N=5$$
etc, with $N=7$ using a $12$-deg equation. I found these using Mathematica's FindRoot command but, unlike the other post, I couldn't find a nice common form for $\gamma$. (The pattern of this family is also different. I had expected $N=7$ to also involve a sextic only.)

Q: Is it true one can find algebraic number $\gamma$ for all prime $N$? What is it for $N=11$? 


Update, Aug 16, 2019
In this comment, Reshetnikov gave the explicit solution to,
$$I\left(\gamma^2;\ \tfrac14,\tfrac14\right) = \tfrac17$$
as,
$$\small\gamma = \frac16\left(5\cos x-\sqrt3\sin x-1-\sqrt3\sqrt{7+4\sqrt7-(11+2\sqrt7)\cos x+\sqrt3(5+2\sqrt7)\sin x}\right)$$
where $x = \tfrac13\arccos\big(\tfrac{13}{14}\big)$.
P.S. I forgot I also found $\gamma$ in this 2016 post as,
$$\gamma = \tfrac12\left(2\cos\tfrac{2\pi}7-\sqrt{2\cos\tfrac{4\pi}7+\sqrt2\csc\tfrac{9\pi}{28}}\right)$$

It seems solving $I(\beta^2;\tfrac{1}{2},\tfrac{1}{4})=\frac1N$ also solves $I(\gamma^2;\tfrac{1}{4},\tfrac{1}{4})=\frac1N$ as they involve reciprocal polynomials. So again following Nemo's lead in this answer, we employ the duplication formula,
$$\frac{1}{2}I(p^2;\tfrac{1}{2},\tfrac{1}{4})=I(1-q^2;\tfrac{1}{2},\tfrac{1}{4})$$ 
where $p,q$ are related by the $8$-deg,
$$p^2 (-1 + 2 q + q^2)^4 = 16 q (-1 + q^2) (1 + q^2)^2\tag1$$
So given a solution to $\displaystyle I(\beta^2;\tfrac{1}{2},\tfrac{1}{4})= \frac{1}{2^n N}$ for $n=0$, the duplication formula enables us to find infinitely many  for $n\geq1$.
For example, the solution to $I(p^2;\tfrac{1}{2},\tfrac{1}{4})=\frac{1}{3}$ is $p = 1-\sqrt3$. One can then solve for $I(\beta^2;\tfrac{1}{2},\tfrac{1}{4})=\frac{1}{6}$ by plugging the known $p$ into $(1)$.
