Evaluate the integral: $\int\limits_{4}^{9} \frac{\sqrt{x}}{(30 - x^{\frac{3}{2}})^2} \, dx$

$\frac{19}{99}$

The correct answer is Option (2) → $\frac{19}{99}$

Let $I = \int\limits_{4}^{9} \frac{\sqrt{x}}{(30 - x^{\frac{3}{2}})^2} \, dx$. We first find the anti derivative of the integrand.

Put $30 - x^{\frac{3}{2}} = t$. Then $-\frac{3}{2} \sqrt{x} \, dx = dt$ or $\sqrt{x} \, dx = -\frac{2}{3} dt$

Thus, $\int \frac{\sqrt{x}}{(30 - x^{\frac{3}{2}})^2} \, dx = -\frac{2}{3} \int \frac{dt}{t^2} = \frac{2}{3} \left[ \frac{1}{t} \right] = \frac{2}{3} \left[ \frac{1}{(30 - x^{\frac{3}{2}})} \right] = F(x) \text{}$

Therefore, by the second fundamental theorem of calculus, we have

$I = F(9) - F(4) = \frac{2}{3} \left[ \frac{1}{(30 - x^{\frac{3}{2}})} \right]_4^9$

$= \frac{2}{3} \left[ \frac{1}{(30 - 27)} - \frac{1}{30 - 8} \right] = \frac{2}{3} \left[ \frac{1}{3} - \frac{1}{22} \right] = \frac{19}{99}$