Evaluate the following integral:
$\int_{4}^{9} \frac{\sqrt{x}}{\left(30-x^{\frac{3}{2}}\right)^{2}} d x$

(A) Let $I = \int_{4}^{9} \frac{\sqrt{x}}{\left(30-x^{\frac{3}{2}}\right)^{2}} d x$.
We use the substitution method. Let $t = 30 - x^{\frac{3}{2}}$.
Then,$dt = -\frac{3}{2} x^{\frac{1}{2}} dx$,which implies $\sqrt{x} dx = -\frac{2}{3} dt$.
When $x = 4$,$t = 30 - 4^{\frac{3}{2}} = 30 - 8 = 22$.
When $x = 9$,$t = 30 - 9^{\frac{3}{2}} = 30 - 27 = 3$.
Substituting these into the integral:
$I = \int_{22}^{3} -\frac{2}{3} \frac{dt}{t^2} = \frac{2}{3} \int_{3}^{22} t^{-2} dt$.
$I = \frac{2}{3} \left[ -\frac{1}{t} \right]_{3}^{22} = \frac{2}{3} \left( -\frac{1}{22} - (-\frac{1}{3}) \right) = \frac{2}{3} \left( \frac{1}{3} - \frac{1}{22} \right)$.
$I = \frac{2}{3} \left( \frac{22 - 3}{66} \right) = \frac{2}{3} \left( \frac{19}{66} \right) = \frac{19}{99}$.