Improper integral of $\int\frac{2}{x^2-1}$

Question

Improper integral of $\int^\infty_3\frac{2}{x^2-1}dx $

I know I need the limit of $\lim_{b \to \infty}$. Solving the integral first:

$$\int\frac{2}{x^2-1}dx = 2 \int\frac{1}{x^2-1}dx = 2\ln|x^2-1|$$

now working on the definite integral

$$(\ln|x^2-1|) ]^b_3 = \ln|b^2-1| - \ln|8|$$

and the limit:

$$\lim_{b \to \infty} (\ln|b^2-1| - \ln|8|) = \infty - 2$$

and them I'm completely lost, because I should have a zero somewhere, to use as a value or as divergent

The answer is $\ln|2|$. Question is on Anton Calculus, 8th Edition, page 576, question 5.

Answer 1

Note that $$2 \int\frac{1}{x^2-1}dx \not= 2ln|x^2-1|.$$

It is: $$2 \int\frac{1}{x^2-1}dx =\int(\frac{1}{x-1}-\frac{1}{x+1})dx =(\ln|x-1|-\ln|x+1|)+C.\,\,$$

Can you proceed from here?

Answer 2

$\int_3^{\infty}\frac{2}{x^2-1}dx=2\int_3^{\infty}\frac{1}{x^2-1}dx=2\int_3^{\infty}\big(\frac{1/2}{x-1}-\frac{1/2}{x+1}\big)dx=\log(\frac{x-1}{x+1})\Big|_3^{\infty}=\cdots $