Question

integral from 2 to infinity of dx/x^2 - 1?

Answer 1

Substitute u = sqrt(x)
du/dx = 1 / (2sqrt(x))
dx = 2sqrt(x) du
dx = 2u du
Inverse function of u: x = sqrt(u)
So:
integral from 2 to infinity of dx/x^2 - 1
\[=\int\limits_{\sqrt{2}}^{\infty} \frac{2u}{u-1}du\]
You can use integration by parts from here.
Note: This is my opinion... Others may not agree