# ln(infinity/infinity)

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..

• Nitpick: All your integrals are lacking "$dx$". – Hans Lundmark Oct 5 '18 at 6:49
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.