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I calculate a integration which results in the below. $$ \log{\left|\frac{x+\sqrt{x^2+1}-1}{x+\sqrt{x^2+1}+1}\right|} $$

I've checked the integration with Wolfram Alpha and it give me this result. $$ \log{\left|\frac{\sqrt{x^2+1}-1}{x}\right|} $$

These expressions seems to have the equal value, but I cannot find the way to transform one to the other.

Is it possible to perform the transformation?
If possible, how can I do that?

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    $\begingroup$ The argument of the first log above is 1. You're missing a sign. $\endgroup$
    – chhro
    Jul 29, 2017 at 12:59
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    $\begingroup$ Apparently you have miscopied something. $\frac{x+ \sqrt{x^2+ 1}- 1}{x+ \sqrt{x^2+ 1}- 1}= 1$ so that first logarithm is 0 for all x, $\endgroup$
    – user247327
    Jul 29, 2017 at 13:02
  • $\begingroup$ I've missed the sign indeed. Thanks! $\endgroup$
    – equal-l2
    Jul 29, 2017 at 13:04

1 Answer 1

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$$ \begin{align} \frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}\cdot\frac{x+1-\sqrt{x^2+1}}{x+1-\sqrt{x^2+1}} &=\frac{(x^2-1)+2\sqrt{x^2+1}-(x^2+1)}{(x+1)^2-(x^2+1)} \\ &= \frac{2\sqrt{x^2+1}-2}{2x} \\ &= \frac{\sqrt{x^2+1}-1}{x} \\ \end{align} $$

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