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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:

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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1 Answer

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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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