Answer
Improper, diverges.
Work Step by Step
The integral has finite bounds so we look at the Definition of Improper Integrals with Infinite Discontinuities.
$f(x)=\displaystyle \frac{1}{(x-3)^{3/2}}$ has an infinite discontinuity at $x=3$, so the case 2 applies:
2. If $f$ is continuous on the interval $(a, b]$ and has an infinite discontinuity at $a$, then$ \displaystyle \int_{a}^{b}f(x)dx=\lim_{c\rightarrow a^{+}}\int_{c}^{b}f(x)dx$.
This is an improper integral.
$\displaystyle \int_{3}^{4}\frac{1}{(x-3)^{3/2}}dx=\lim_{c\rightarrow 3^{+}}\int_{c}^{4}(x-3)^{-3/2}dx$
$=\displaystyle \lim_{c\rightarrow 3^{+}}[\frac{1}{-1/2}(x-3)^{-1/2}]_{c}^{4}$
$=\displaystyle \lim_{c\rightarrow 3^{+}}[-2(x-3)^{-1/2}]_{c}^{4}$
$=-2\displaystyle \lim_{c\rightarrow 3^{+}}[1-\frac{1}{\sqrt{c-3}}]$
$($the second term diverges, as the denominator $\rightarrow$ 0)
$=\infty$
Diverges.
