Step-by-Step Solution
Key Concept: General
Put $4x = t \Rightarrow \int \frac{dx}{\sqrt{9 - 16x^2}} = \int \frac{dt}{4\sqrt{9 - t^2}} = \frac{1}{4} \sin^{-1} \frac{t}{3} + C = \frac{1}{4} \sin^{-1} \frac{4}{3}x + C$
Correct Answer: $\frac{1}{4} \sin^{-1} \frac{4}{3}x + C$