# How do you find the integral of int 1/ (x-2)^2 dx  from 0 to infinity?

${\int}_{0}^{\infty} \frac{1}{x - 2} ^ 2 \mathrm{dx} = {\int}_{0}^{\infty} {\left(x - 2\right)}^{- 2} \mathrm{dx} = {\left[{\left(x - 1\right)}^{- 1} / \left(- 1\right)\right]}_{0}^{\infty}$
=${\left[\frac{1}{x - 1}\right]}_{\infty}^{0} = \frac{1}{0 - 1} - \frac{1}{\infty - 1} = - 1 - \frac{1}{\infty} = - 1$