Answer
Improper, diverges.
Work Step by Step
$f(x)=\displaystyle \frac{1}{(x-3)^{3/2}}$ has an infinite discontinuity at $x=1\in[0,2]$ so
case 3 of Improper Integrals with Infinite Discontinuities applies.
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 $\displaystyle \int_{a}^{b}f(x)dx=\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx$.
$\displaystyle \int_{0}^{2}\frac{1}{(x-1)^{2}}dx=\lim_{b\rightarrow 1^{-}}\int_{0}^{b}\frac{1}{(x-1)^{2}}d\mathrm{x}+\lim_{c\rightarrow 1^{+}}\int_{c}^{2}\frac{1}{(x-1)^{2}}dx$
$\displaystyle \left[\begin{array}{ll}
u=\frac{1}{x-1} & \\
du=-\frac{1}{(x-1)^{2}}dx &
\end{array}\right],\quad\int\frac{1}{(x-1)^{2}}dx=\int-du.$
$\displaystyle \int_{0}^{2}\frac{1}{(x-1)^{2}}dx=\lim_{b\rightarrow 1^{-}}[-\frac{1}{x-1}]_{0}^{b}+\lim_{c\rightarrow 1^{+}}[-\frac{1}{x-1}]_{c}^{2}$
$=\displaystyle \lim_{b\rightarrow 1^{-}}[-\frac{1}{b-1}-1] +\lim_{c\rightarrow 1^{+}}[-\frac{1}{2-1}+\frac{1}{c-1}] $
$=(\infty-1)+(-1+\infty)$
Diverges.
