How to prove $\int_0^1\frac{1-x}{(\ln x)(1+x)}\ dx=\ln\left(\frac2{\pi}\right)$?

We've $\displaystyle \sum_{n \ge 1} (-1)^n \left[\log(n+1)- \log(n)\right] = \sum_{n \ge 1}(-1)^n\log\left(\frac{n+1}{n}\right) = \log(\mathcal{P})$ where:

$\displaystyle \mathcal{P} = \prod_{n \ge 1}\bigg(\frac{n+1}{n}\bigg)^{(-1)^n} = \prod_{n \ge 1} \bigg(\frac{2n+1}{2n}\bigg) \bigg(\frac{2n}{2n-1}\bigg)^{-1} = \prod_{n \ge 1} \bigg(\frac{2n+1}{2n}\cdot \frac{2n-1}{2n}\bigg) = \frac{2}{\pi}. $

Where the last step is the Wallis product for $\pi$.

Define $I(a) := \int_0^1 \frac{x^a(1-x)}{\ln(x)(1+x)}dx$. Then, for $a > -1$, $I'(a) = \int_0^1 \frac{x^a(1-x)}{1+x}dx = \frac{2_2F_1(1,1+a;2+a;-1)-1}{1+a}$, so $I(a) = 2\log \left(\Gamma(\frac{a}{2}+1)\right)-2\log\left(\Gamma(\frac{a+1}{2})\right)-\log(a+1)+\log(2)$. Plug in $a=0$.