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

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).

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