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I'm looking for the solution to the integral $$\int_0^1 \left(\frac{\ln(x)}{1-x}\right)^2dx$$ I solved and know that the solution to $$-\int_0^1 \frac{\ln(x)}{1-x}dx = \frac{\pi^2}{6}$$ through a taylor series argument, and am wondering if a similar approach is the best way to go.

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marked as duplicate by Martin R, Community Dec 20 '18 at 19:38

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hints

By parts

$$u=\frac{-1}{1-x}$$ $$v=(\ln(x))^2$$

the integral becomes

$$[uv]+2\int\frac{\ln(x)}{x(1-x)}$$

with $$\frac{1}{x(1-x)}=\frac 1x+\frac{1}{1-x}$$

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