Proving that $\int_0^\infty\Big(\sqrt[n]{1+x^n}-x\Big)dx~=~\frac12\cdot{-1/n\choose+1/n}^{-1}$ 
How can we prove, without employing the aid of residues or various transforms, that, for $n>2$ $$\int_0^\infty\Big(\sqrt[n]{1+x^n}-x\Big)dx~=~\frac12\cdot{-1/n\choose+1/n}^{-1}$$


Motivation: In my previous question, thanks to Will Jagy's simple but brilliant answer, I was able to express the area of the superellipse $x^n+y^n=r^n$, for odd values of $n=2k+1$, with $k\in\mathbb N^*$, as 
$A_n=r^2\displaystyle\cdot{2/n\choose1/n}^{-1}+r^2\cdot{-1/n\choose+1/n}^{-1}$, where the first term, representing the surface inside the 
first sector or quadrant, comes from a simple evaluation of the beta function.

My fruitless efforts and endeavors:
$\big(1\big).$ Breaking up the integration interval into $(0,1)$ and $(1,\infty)$, expanding the integrand into an appropriate binomial series for each of the two cases, and then switching the order of summation and integration. In the end, this made me rewrite the initial question in terms of an infinite series, but led me no closer to finding an answer:
$$2a\sum_{k=1}^\infty{a\choose k}\Bigg(\frac1{k+a}+\frac1{k-2a}\Bigg)={-a\choose+a}^{-1}-1.$$

$\big(2\big).$ Letting $x=\sqrt[n]{\sinh^2t~}$. Needless to say, that did not get me very far either.
$\big(3\big).$ Letting $x^n=u$, we are able to obtain an expression in terms of incomplete beta functions.
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$\ds{\int_{0}^{\infty}\pars{\root[n]{1 + x^{n}} - x}\,\dd x
     =\half\,{-1/n \choose 1/n}^{-1}:\ {\large ?}.\qquad\qquad n > 2}$.

\begin{align}&\overbrace{\color{#c00000}{
\int_{0}^{\infty}\pars{\root[n]{1 + x^{n}} - x}\,\dd x}}
^{\ds{x^{n}\ \mapsto x}}\ =\
\int_{0}^{\infty}\bracks{\pars{1 + x}^{1/n} - x^{1/n}}\,{1 \over n}\,x^{1/n - 1}
\,\dd x
\\[3mm]&={1 \over n}\ \overbrace{\int_{0}^{\infty}\bracks{
\pars{1 + x}^{1/n}x^{1/n - 1} - x^{2/n - 1}}\,\dd x}
^{\ds{t\ \equiv {1 \over 1 + x}\ \imp\ x = {1 \over t} - 1}}
\\[3mm]&={1 \over n}\int_{1}^{0}\bracks{
t^{-1/n}\pars{{1 \over t} - 1}^{1/n - 1}
-\pars{{1 \over t} - 1}^{2/n - 1}}\,\pars{-\,{\dd t \over t^{2}}}
\\[3mm]&={1 \over n}\int_{0}^{1}\bracks{
t^{-2/n - 1}\pars{1 - t}^{1/n - 1} - t^{-2/n - 1}\pars{1 - t}^{2/n - 1}}\,\dd t
\end{align}

\begin{align}&\color{#c00000}{
\int_{0}^{\infty}\pars{\root[n]{1 + x^{n}} - x}\,\dd x}
={1 \over n}\int_{0}^{1}{t^{1/n - 1} - 1 \over \pars{1 - t}^{2/n + 1}}\,\dd t-
{1 \over n}\int_{0}^{1}{t^{2/n - 1} - 1 \over \pars{1 - t}^{2/n + 1}}\,\dd t
\end{align}
Both integrals converge when $\ds{\Re\pars{n} > 2}$. Moreover,
$$
\int_{0}^{1}{t^{a} - 1 \over \pars{1 - t}^{2/n + 1}}\,\dd t
=\half\,n + {a!\pars{-2/n - 1}! \over \pars{a - 2/n}!}
$$

such that
  \begin{align}&\left.\color{#66f}{\large
\int_{0}^{\infty}\pars{\root[n]{1 + x^{n}} - x}\,\dd x}\,\right\vert_{\ n\ >\ 2}
={1 \over n}\,{\pars{1/n - 1}!\pars{-2/n - 1}! \over \pars{-1/n - 1}!}
\\[3mm]&={1 \over n}\,
{\bracks{n\pars{1/n}!}\bracks{\pars{-n/2}\pars{-2/n}!} \over -n\pars{-1/n}!}
=\half\bracks{\pars{-1/n}! \over \pars{1/n}!\pars{-2/n}!}^{-1}
=\color{#66f}{\large\half\,{-1/n \choose 1/n}^{-1}}
\end{align}

