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I found it from a exercise book that the following integral $$ \int_{0}^{1} \frac{\log(\frac{1-x}{1+x})}{x\sqrt{1-x^2}}dx $$ can be calculated by certain substitution: $$ \int_{0}^{1} \frac{\log(\frac{1-x}{1+x})}{x\sqrt{1-x^2}}\,dx =4\int_{0}^{1} \frac{\log (y)}{1-y^2} dy\tag{1} $$ The book does not provides any details what substitution is used.

Question: What substitution should I use to get (1)?

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  • $\begingroup$ Ya,it was a typo,thanka for figuring out. $\endgroup$ – NewBornMATH Oct 14 at 14:51
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Note that there is a missing factor of $4$.

The substitution is pretty much there as the argument of the logarithm:$$\frac{1-x}{1+x}=t\Rightarrow x=\frac{1-t}{1+t}\Rightarrow dx=-\frac{2}{(1+t)^2}dt$$ $$\Rightarrow \int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)}{x\sqrt{1-x^2}}dx=\int_0^1 \frac{\ln t}{\sqrt t (1-t)}dt\overset{t=y^2}=4\int_0^1 \frac{\ln y}{1-y^2}dy$$ Here is a way to solve the last integral in case one's interested.

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