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I am taking AP Calculus BC this year, and we are going over improper integrals. I was just doing this integral, and was wondering what exactly the ln(inf/inf) is. Here is my work:enter image description here

I believe my teacher had said something about how it equals 1 because x-1 and x+1 are of the same degree. Does that make sense? Does L'Hôpital's rule have anything to do with it? I'm pretty sure that's only for limits, but who knows..

Let me know your thoughts!

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  • $\begingroup$ Nitpick: All your integrals are lacking "$dx$". $\endgroup$ – Hans Lundmark Oct 5 '18 at 6:49
  • $\begingroup$ @HansLundmark ah yes, my bad. If I remember correctly, the 'dx' is the width of each Riemenn area, if you think about it in terms of that. Is that right? $\endgroup$ – Addison Oct 5 '18 at 13:02
  • $\begingroup$ Yes, that's right. And that's why the integrals look a bit funny without it. $\endgroup$ – Hans Lundmark Oct 5 '18 at 13:13
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The antiderivative is right. But you can’t “plug in $\infty$”: you need to compute a limit: $$ \lim_{x\to\infty}\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|= \lim_{x\to\infty}\frac{1}{2}\ln\left|\frac{1-\frac{1}{x}}{1+\frac{1}{x}}\right|= \frac{1}{2}\ln 1=0 $$ Similarly for the lower bound: $$ \lim_{x\to1^+}\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|=-\infty $$ The integral is not convergent.

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  • $\begingroup$ I see, so infinity cannot be treated as a number. It must be treated as a limit, correct? $\endgroup$ – Addison Oct 4 '18 at 23:54
  • $\begingroup$ @Addison: correct $\endgroup$ – Clayton Oct 5 '18 at 0:05

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