Evaluating $~\int_0^1\sqrt{\frac{1+x^n}{1-x^n}}~dx~$ and $~\int_0^1\sqrt[n]{\frac{1+x^2}{1-x^2}}~dx$ How could we prove that

$$\int_0^1\sqrt{\frac{1+x^n}{1-x^n}}~dx~=~a\cdot2^{a-1}~\bigg[\frac12~B\bigg(\frac a2,~\frac a2\bigg)~+~B\bigg(\dfrac{a+1}2,~\dfrac{a+1}2\bigg)\bigg],$$

where $a=+~\dfrac1n$ , and

$$\int_0^1\sqrt[n]{\frac{1+x^2}{1-x^2}}~dx~=~a\cdot2^{a-1}~\bigg[\frac12~B\bigg(\frac a2,~\frac a2\bigg)~-~B\bigg(\dfrac{a+1}2,~\dfrac{a+1}2\bigg)\bigg],$$

where $a=-~\dfrac1n$ ?

This question arose as a generalization of the fact that the arc length of the $($co$)$sine function over an interval of the form $\bigg(k~\dfrac\pi2,~m~\dfrac\pi2\bigg)$, with $k,~m\in\mathbb Z$, can be expressed in terms of $\Gamma$ functions, which was somewhat surprising, since the evaluation of such arc lengths usually involves elliptic integrals; indeed, it was one of the main historical reasons for defining them in the first place.

I tried substituting $x^n=\cos2t$, and then employing the well-known trigonometric formulas for $1\pm\cos2t$, in the hopes of reducing the original problem to a beta function by means of Wallis' integrals, since $\sqrt{\dfrac{1+x^n}{1-x^n}}~=~\cot(t)$, but this approach ultimately lead me nowhere $($which is not to imply that the same method might not prove fruitful in someone else's skilled hands$)$.
 A: First integral:
Let us multiply the numerator and denominator of the integrand by $\sqrt{1+x^n}$ , so that the initial integral becomes
$$\mathcal{I}=\underbrace{\int_0^1\frac{dx}{\sqrt{1-x^{2n}}}}_{\mathcal{I}_1}+
\underbrace{\int_0^1\frac{x^ndx}{\sqrt{1-x^{2n}}}}_{\mathcal{I}_2}.$$
After the substitution $x=\sin^{\frac1n}\theta$ the integrals $\mathcal{I}_{1,2}$ reduce to integral representations of two beta functions:
$$\mathcal{I}_1=\frac{1}{2n}B\left(\frac12,\frac1{2n}\right),\qquad 
\mathcal{I}_2=\frac{1}{2n}B\left(\frac12,\frac{1+n}{2n}\right).$$ 
It can be easily shown (using duplication formula for gamma function or repeating the computation from this link) that 
\begin{align*}
B\left(\frac12,\frac1{2n}\right)&=2^{\frac1n-1}B\left(\frac1{2n},\frac1{2n}\right),\\
B\left(\frac12,\frac{1+n}{2n}\right)&=
2^{\frac1n}B\left(\frac{1+n}{2n},\frac{1+n}{2n}\right),
\end{align*}
which proves the first formula of the question. Hopefully this is not too mysterious.

Second integral:
The change of variables
$$y=\sqrt[n]{\frac{1-x^2}{1+x^2}} \qquad \Longleftrightarrow\qquad x=\sqrt{\frac{1-y^n}{1+y^n}}$$
transforms the integral into
$$\int_0^1\frac{ny^{n-2}dy}{\left(1+y^n\right)\sqrt{1-y^{2n}}}
=\int_0^1\frac{ny^{n-2}\left(1-y^n\right)dy}{\left(1-y^{2n}\right)^{\frac32}}=\int_0^1\frac{\left(1-y^n\right)}{y}d\left(\frac{y^n}{\sqrt{1-y^{2n}}}\right).$$
Now it suffices to integrate by parts and apply exactly the same procedure (two beta functions at the first step and duplication formula at the second one).
A: Ok maybe not the most elegant solution but nevertheless:
I concentrate on the integral first integral and call it $I_1$. Using a substitution $x^n=\cos^2(y),\,dx=-\frac{2}{n}\sin(x)\cos^{\frac{2}{n}-1}(x)$ we obtain
$$
I_1=\frac{2}{n}\int_{0}^{\pi/2}\cos^{\frac{2}{n}-1}(y)\sqrt{1+\cos^2(y)}
$$
Expanding the square root as a Taylor series allows us to write after exchanging summation and integration:
$$
I_1=\frac{2}{n}\sum_{m=0}^{\infty}\binom{1/2}{m}\int_0^{\pi/2}\cos^{2m+\frac{2}{n}-1}(y)
$$
The integral is now in the form of Beta function and we obtain
$$
I_1=\frac{\sqrt{\pi}}{n}\sum_{m=0}^{\infty}\binom{1/2}{m}\frac{\Gamma \left(m+\frac{1}{n}\right) }{\Gamma \left(m+\frac{1}{n}+\frac{1}{2}\right)}
$$
Remembering $\binom{1/2}{m}=\frac{(-1)^m\Gamma(2m+1)}{\Gamma(m+1)^2}\frac{1}{(1-2m)4^m}$
And using some basic properties of the Gamma function (duplication formula and the reduction formula for $\Gamma(\frac{1}{2}+n)$)this boils down to 
$$
I_1=\frac{1}{n}\sum_{m=0}^{\infty}(-1)^m\frac{\Gamma \left(m-\frac{1}{2}\right)}{m!}\frac{\Gamma \left(m+\frac{1}{n}\right) }{\Gamma \left(m+\frac{1}{n}+\frac{1}{2}\right)}
$$
Now using the the defintion of the Hypergeometric function together with the defintion of Pochhammer symbol we may rewrite this as
$$
I_1=\frac{1}{n}B(\frac{1}{2},\frac1n)_2F_1\left(-\frac{1}{2},\frac{1}{n},\frac{1}{2}+\frac{1}{n}\right)
$$
Now using Eulers Transformation this yields
$$
I_1=\frac{1}{2n}B(\frac{1}{2},\frac1n)_2F_1\left(1+\frac{1}{n},\frac{1}{2},\frac{1}{2}+\frac{1}{n},-1\right)
$$
Which may accoording to this site can be evaluated in terms of $\Gamma$-functions yielding the correct result.
Remarks:
I'm not very satisfied with this answer because it contains (what i really hate to be honest) to much Hypergeometric hocus-pocus. I will try to add a proof to the crucial identity tomorrow, but hopefully we find a more elegant way for this problem...
Edit:
For a proof of the above mentioned identity have a look 
here,chapter 2
