Evaluate the definite integral \( \int_{0}^{2/3} \frac{dx}{4+9x^2} \).

Rewriting as \( \frac{1}{9} \int_{0}^{2/3} \frac{dx}{\frac{4}{9}+x^2} \), integration gives \( \frac{1}{9} \cdot \frac{3}{2} [\tan^{-1}(\frac{3x}{2})]_0^{2/3} = \frac{1}{6} \cdot \frac{\pi}{4} = \frac{\pi}{24} \).