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How to evaluate $$\int_0^1 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}} dx$$

I am clueless, I doubt if this has a closed form either.

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Sub $u=(1+x)/(1-x)$; the resulting integral is

$$\int_1^{\infty} du \frac{\log{u}}{(u-1) \sqrt{u}} = 4 \int_1^{\infty} dv \frac{\log{v}}{v^2-1} = 2 \int_0^{\infty} dv \frac{\log{v}}{v^2-1}$$

The latter integral is easily attacked using contour integration methods, as done in this answer. The result is $\pi^2/2$. Alternatively, you may express the integral over the interval $[0,1]$, Taylor expand the denominator, and obtain a familiar sum.

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