TITLE: Evaluating $\int_{0}^{1} \frac{dx}{1+{}_2F_{1}\left(\frac{1}{n},x;\frac{1}{n};\frac{1}{n}\right)}$
QUESTION [8 upvotes]: On a lark (as a followup to this question), I was playing around with Wolfram alpha, and it seems that
$$\int_{0}^{1} \frac{dx}{1+{}_2F_{1}\left(\frac{1}{n},x;\frac{1}{n};\frac{1}{n}\right)} = \frac{\log\left(\frac{2n}{2n-1}\right)}{\log\left(\frac{n}{n-1}\right)}$$
Could someone take a stab at proving this?

REPLY [3 votes]: To make GEdgar's solution more explicit: whenever you have equal numerator and denominator parameters in a hypergeometric function, you can cancel them:
$${}_2 F_1\left({{1/n}\atop{}}{{}\atop{1/n}}{{x}\atop{}}\mid \frac1{n}\right) = \sum_{j=0}^\infty \frac{(1/n)_j (x)_j}{(1/n)_j}\frac{(1/n)^j}{j!} = \sum_{j=0}^\infty (x)_j\frac{(1/n)^j}{j!} = {}_1 F_0\left({{x}\atop{—}}\mid \frac1{n}\right)$$
${}_1 F_0$ can subsequently be re-expressed as a binomial series by exploiting identities for converting Pochhammer symbols to binomial coefficients...