# How to prove $\int_{-1}^{1}{\mathrm dx\over x}\cdot\sqrt{1+x\over 1-x}\cdot{2x(x-1)+\ln((1-x)(1+x)^3)\over x}=\pi^2-4\pi?$

Motivated by this interesting question and ideas came from lemma 2.

Then we have this integral:

$$\int_{-1}^{1}{\mathrm dx\over x}\cdot\sqrt{1+x\over 1-x}\cdot{2x(x-1)+\ln((1-x)(1+x)^3)\over x}=\pi^2-4\pi.\tag1$$

Here is my try:

Split out $(1)$ then we have

$$2\int_{-1}^{1}{\mathrm dx\over x}\cdot\sqrt{1-x^2}+\int_{-1}^{1}{\mathrm dx\over x^2}\cdot\sqrt{1+x\over 1-x}\cdot\ln((1-x)(1+x)^3)=I_1+I_2.\tag2$$

For $I_1$, let $x=\sin u$ then we have

$$I_1=\int_{-\pi/2}^{\pi/2}\mathrm du{\cos^2(u)\over \sin(u)},\tag3$$ as for $I_2$ seems too complicate to make an attempt. Probably easy but I can't see it.

How to prove (1) and (2)?

• Your $I_1$ diverges. – xpaul Apr 27 '17 at 21:58
