Question

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

Answer 1

-1

Explanation:

`int_0^oo1/(x-2)^2dx=int_0^oo(x-2)^(-2)dx=[(x-1)^(-1)/(-1)]_0^oo`
=`[1/(x-1)]_oo^0=1/(0-1)-1/(oo-1)=-1-1/oo=-1`