I disagree with correct solution in the OP and will present my own analysis for your scrutiny. First, let's reduce the clutter and set $a=E_n$. We seek to find $a$ such that
$$2\int_0^{a^{1/4}}\sqrt{a-x^4}dx=\left(n+\frac{1}{2}\right)$$
Now let $x^4=at$ or $x=(at)^{1/4}$, then
$$
dx=\frac{a^{1/4}}{4}t^{-3/4}dt\\
\sqrt{a-x^4}=\sqrt{a}\sqrt{1-t}\\
x=a^{1/4}\to t=1
$$
Substituting and rearranging we get
$$\frac{a^{3/4}}{2}\int_0^1 t^{-3/4}(1-t)^{1/2}dt=\left(n+\frac{1}{2}\right)$$
Introducing the complete beta function,
$$B(\nu,\mu)=\int_0^1 t^{\nu-1}(1-t)^{\mu-1}dt=\frac{\Gamma(\nu)\Gamma(\mu)}{\Gamma(\nu+\mu)}$$
Clearly, $\nu=1/4$ and $\mu=3/2$ and we can then show that
$$a=\left[ \frac{2\left(n+\frac{1}{2}\right)\Gamma(7/4)}{\Gamma(1/4)\Gamma(3/2)}\right]^{4/3}$$
At the OP's suggestion, we can substitute
$$
\Gamma(7/4)=(3/4)\Gamma(3/4)\\
\Gamma(3/2)=\sqrt{\pi}/2\\
\Gamma(1/4)\Gamma(3/4)=\pi\sqrt{2}
$$
and demonstrate that
$$E_n=\left[ \frac{2\Gamma(3/4)^2\left(n+\frac{1}{2}\right)}{\pi\sqrt{2\pi}}\right]^{4/3}$$
So, it would appear that the original correct solution was missing a factor of $\pi$ in the denominator.
The solution I present here has been validated numerically for $n\in\mathbb{R^+}$, i.e., not just $n\in\mathbb{Z}$.
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\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}\bracks{1 - \pars{1 - x^{3}}^{\root{2}}}^{\root{3}}x^{2}\,\dd x}
\\[5mm] \stackrel{x^{3}\ \mapsto\ y}{=}\,\,\,&
{1 \over 3}\int_{0}^{1}
\bracks{1 - \pars{1 - y}^{\root{2}}}^{\root{3}}\dd y
\\[5mm]
\stackrel{1 - y\ \mapsto\ z}{=}\,\,\,&
{1 \over 3}\int_{0}^{1}
\bracks{1 - z^{\root{2}}}^{\root{3}}\dd z
\\[5mm] \,\,\,\stackrel{\pars{1 - z}^{\root{2}}\ \mapsto\ t}{=}\,\,\,&
{\root{2} \over 6}\int_{0}^{1}t^{\root{3}}\pars{1 - t}^{\root{2}/2 - 1}
\,\dd t
\\[5mm] = &\
\bbx{{\root{2} \over 6}\,\mrm{B}\pars{\root{3} + 1,{\root{2} \over 2}}}
\approx 0.1546
\end{align}
$\ds{\mrm{B}}$ is the
Beta Function.
Best Answer
For the second integral, make the substitution $ \displaystyle x = \frac{u}{1+u}$.
Then
$$ \begin{align} 2 \int_{0}^{1} \frac{dx}{(2-x) \sqrt[3]{x^2(1-x)}} &= 2 \int_{0}^{\infty} \frac{du}{(2+u)u^{2/3}} \\ &= \int_{0}^{\infty} \frac{du}{\left(1+ \frac{u}{2}\right)u^{2/3}} \\ &= \frac{2}{2^{2/3}} \int_{0}^{\infty} \frac{1}{(1+t)t^{2/3}} \, dt \\ &= \frac{2}{2^{2/3}} \int_{0}^{\infty} \frac{t^{1/3-1}}{(1+t)^{1/3+2/3}} \, dt \\ &= \frac{2}{2^{2/3}} B \left(\frac{1}{3}, \frac{2}{3} \right) \tag{1} \\&= \frac{2}{2^{2/3}} \frac{\Gamma \left(\frac{1}{3} \right) \Gamma \left(1- \frac{1}{3} \right)}{\Gamma(1)} \\ &= \frac{2}{2^{2/3}} \frac{2 \pi}{\sqrt{3}} \tag{2} \\ &= \frac{2^{4/3} \pi}{\sqrt{3}}. \end{align}$$
$(1)$ https://en.wikipedia.org/wiki/Beta_function#Properties
$(2)$ https://en.wikipedia.org/wiki/Gamma_function#General