# How substitution can I use to get $\int_{0}^{1} \frac{\log(\frac{1-x}{1+x})}{x\sqrt{1-x^2}}dx=4\int_{0}^{1} \frac{\log (y)}{1-y^2} dy$?

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)?

• Ya,it was a typo,thanka for figuring out. – NewBornMATH Oct 14 at 14:51

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.