Note that if the integral:
\displaystyle \int_0^3 \frac 1 {(1 - x)^2} \mathrm dx
is to converge in the usual sense then:
\displaystyle \int_0^a \frac 1 {(1 - x)^2} \mathrm dx
should be finite for 0 \le a < 3 too.
However note that:
\displaystyle \int_0^1 \frac 1 {(1 - x)^2} \mathrm dx = \left[\frac 1 {1 - x}\right]_0^1 = \lim_{x \to 1^-} \left(\frac 1 {1 - x} - 1\right) = \infty
so the integral in question diverges, and so the argument starts on a faulty premise. It may be enough just to point out that the integrand is unbounded as x \to 1 so the integral should diverge, but this is a fairly standard check.
(it should, in any case, be alarming that the integral of a positive function is apparently negative)
Annendum: Worse - from an analytic perspective
The fundamental theorem of calculus is used here errorneously. More specifically, FTC applies to integrals of continuous functions, but the integrand here, \dfrac 1 {(1 - x)^2} is not continuous on [0,3], (it’s not continuous on (0,3) so cannot be understood as an improper integral either) and cannot be extended to a continuous function on [0,3] since it approaches no finite limit as x \to 1. Discontinuous functions can still be integrable however, if they are bounded and don’t have many singularities. (the set of singularities should have measure zero, being countable or finite suffices)
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