Calculus of a Single Variable, Chapter 8, 8.8, Section 8.8, Problem 11

Integral is improper if we have to take limit in order to calculate it. This can happen if we have infinite values of integration or if the interval if integration contains point(s) where the function is not defined. The latter is the case here because the function we are integrating is not defined for x=1.
Because this point is within the interior of the interval of integration (not at the endpoint) we first must write this integral as a sum of two integrals.
int_0^2 1/(x-1)^2 dx=int_0^1 1/(x-1)^2dx+int_1^2 1/(x-1)^2dx=
Substitute u=x-1 => du=dx. Since u=x-1 all bounds of integration become lower by 1.
int_-1^0 1/u^2 du+int_0^1 1/u^2 du=-1/u|_-1^0-1/u|_0^1=
lim_(u to 0^-) -1/u+1/-1-1/1+lim_(u to 0^+) 1/u=
Notice the use of directional limits (from the left for the first and from the right for the second integral).
-(-infty)-2+infty=infty
As we can see the integral diverges.
The image below shows the graph of the function and area under it corresponding to the value of the integral. Asymptotes of the graph are x-axis and line x=1. The other image (the red one) shows the graph of the function f(u)=1/u. There we see why the two directional limits have different values.