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Evaluate the improper integral $\int\limits_0^1\left(\frac1{\log x} + \frac1{1-x}\right)^2 dx = \log2\pi - \frac12 = 0.33787...$

With integration by parts we get from $\int\limits_0^1\left(\frac1{\log x} + \frac1{1-x}\right) dx = \gamma$

the similar integral $\int\limits_0^1\left(\frac1{\log^2 x} - \frac{x}{(1-x)^2}\right)dx = \gamma-\frac12$

But we need $\int\limits_0^1\left(\frac2{(1-x)\log x} + \frac{1+x}{(1-x)^2}\right)dx = 0.260661401507813...$

to get the integral in question. In question series from one of Coffey's papers involving digamma, $\gamma$, and binomial there is a hint of connection to Stieltjes constants.

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Most of my ideas on this come from using $(\ln \ln(1-x))' = \frac{-1}{(1-x)\ln(1-x)}$.

From there, there is some tidying up to do and integration by parts, but I think on the whole it is fairly doable.

To elaborate,

$\frac{2}{x\ln(1-x)} = \frac{1-x}{x}\frac{2}{(1-x)\ln(1-x)} = \frac{2}{(1-x)\ln(1-x)} - \frac{2x}{(1-x)\ln(1-x)}$.

We would use by parts on integrating the second expression.

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