Integral $\int_{-1}^{1}\frac{\ln(2x^2+2x+1)}{x\sqrt{1-x^2}} \mathrm dx$

Asked Apr 04 '23 at 19:53

Active Apr 04 '23 at 19:53

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I was trying to solve one of the famous integrals on this site:

$$I=\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2x^2+2x+1}{2x^2-2x+1}\right) \mathrm dx$$

I was able to reduce it to the form:

$$I=2\int_{-1}^{1}\frac{\ln(2x^2+2x+1)}{x\sqrt{1-x^2}} \mathrm dx$$

I have verified that both the integrals are matching.
But I am unable to figure it out after this.
In the original post I didn't see a similar approach to this.

asked Apr 04 '23 at 19:53

Miracle Invoker

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This answer to the post in question is essentially what you have. Rewrite $\tanh^{-1}$ in terms of logarithms and simplify to get your integral. – user170231 Apr 04 '23 at 20:02
@user170231 Thank you for the link, I seem to have missed that answer as I haven't dealt with inverse hyperbolic trigonometric functions. – Miracle Invoker Apr 05 '23 at 07:31

Integral $\int_{-1}^{1}\frac{\ln(2x^2+2x+1)}{x\sqrt{1-x^2}} \mathrm dx$

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