$\int_{2}^{\infty}\frac{dx}{x^2-1}$ converge or diverge How can I find whether the integral converge or diverge.
I did the following.
$\int_{2}^{\infty}\frac{dx}{x^2-1}$
I did the following
$\frac{a}{x-1}+\frac{b}{x+1}$
$A(x+1)+B(x-1)=1$
$t\rightarrow\infty$
$\int_{2}^{t}\frac{1/2}{x-1}-\frac{1/2}{x+1}$
$\frac{1}{2}[\ln(x-1)-\ln(x+1)$
$\ln(t-1)-ln(t+1)-\ln(2-1)-\ln(2+1)$
Then I got
$\infty-\infty-ln(1)-ln(2)$
Since I got an intermediate form I did the hospital rule.
$\frac{\ln(t+1)}{1/\ln (t-1)}$
then I took the derivative
$\frac{\frac{1}{t+1}}{\frac{1}{1/t-1}}$
Thus I got
$\frac{1}{(t-1)(t+1)}$ as t approach infity zero
thus it is convergent at at $ln(1)-ln(3)$

One can prove convergence, without evaluating, by noting that for $x\ge 2$, we have $x^2-1\gt x^2-\frac{x^2}{2}=\frac{x^2}{2}$. Thus
$$0\lt \frac{1}{x^2-1}\lt \frac{2}{x^2}.$$
It is a standard fact that $\int_2^\infty \frac{1}{x^2}\,dx$ converges, so our integral does. 
