Answer
The integral was evaluated as if the integrand is continuous on the interval.
It is not, so the integral is improper, and diverges.
Using a CAS, the result is "?" (undefined).
Work Step by Step
$\displaystyle \int_{-2}^{2}\frac{-2}{(x-1)^{3}}dx$ is an improper integral, because $\displaystyle \frac{-2}{(x-1)^{3}}$ is undefined at $x=1\in[-2,2].$
From the Definition of Improper Integrals with Infinite Discontinuities,
3. If $f$ is continuous on the interval $[a, b]$, except for some $c$ in $(a, b)$ at which $f$ has an infinite discontinuity, then
$\displaystyle \int_{a}^{b}f(x)dx=\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx$
$\displaystyle =\lim_{c\rightarrow b^{-}}\int_{a}^{c}f(x)dx+\lim_{c\rightarrow a^{+}}\int_{c}^{b}f(x)dx$.
$\displaystyle \int_{-2}^{2}\frac{1}{(x-1)^{3}}dx=\lim_{c\rightarrow 1^{-}}\int_{-2}^{c}-2(x-1)^{-3}dx+\lim_{c\rightarrow 1^{+}}\int_{c}^{2}-2(x-1)^{-3}dx$.
$=\displaystyle \lim_{c\rightarrow 1^{-}}[\frac{1}{(x-1)^{2}}]_{-2}^{c}+\lim_{c\rightarrow 1^{+}}[\frac{1}{(x-1)^{2}}]_{c}^{2}$
$=(\displaystyle \infty-\frac{1}{9})+(1-\infty)$
Indefinite form $\infty-\infty$,
the integral is undefined.
The integral was evaluated as if the integrand is continuous on the interval.
It is not, so the integral is improper, and diverges.
Using a CAS, the result is "?" (undefined).
