Step 1: Use a substitution instead of memorising the formula:
Let $3x = 2\tan\theta$, so $9x^2 = 4\tan^2\theta$ and $3\,dx = 2\sec^2\theta\,d\theta$.
Step 2: Rewrite the integral in terms of $\theta$:
$4 + 9x^2 = 4(1+\tan^2\theta) = 4\sec^2\theta$, so the integrand becomes $\dfrac{(2/3)\sec^2\theta\,d\theta}{4\sec^2\theta} = \dfrac16 d\theta$.
Step 3: Change the limits and integrate:
At $x=0$, $\theta=0$; at $x=2/3$, $\tan\theta=1$, so $\theta=\pi/4$. The integral becomes $\displaystyle\int_0^{\pi/4}\frac16\,d\theta = \frac16\cdot\frac{\pi}{4} = \frac{\pi}{24}$.
Final Answer:
The definite integral equals $\pi/24$.
\[ \boxed{\dfrac{\pi}{24}} \]