$\displaystyle\int_0^{2/3} \dfrac{dx}{4+9x^2}$ equals

1. Factor: $4+9x^2=9\left(x^2+\tfrac{4}{9}\right)$, with $a=\tfrac{2}{3}$. 2. $\int \dfrac{dx}{9(x^2+a^2)}=\dfrac{1}{9}\cdot\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}=\dfrac{1}{6}\tan^{-1}\dfrac{3x}{2}$. 3. Evaluate $0$ to $\tfrac{2}{3}$: $\dfrac{1}{6}\left(\tan^{-1}1-0\right)=\dfrac{1}{6}\cdot\dfrac{\pi}{4}$. 4. $=\dfrac{\pi}{24}$. _Source: syllabus Mathematics Ch 7 "Integrals", p.338_