Show that

$\int \frac{d x}{4+9 x^{2}}=\int \frac{d x}{(2)^{2}+(3 x)^{2}}$

Put $3 x=t \Rightarrow 3 d x=d t$

$\therefore \int \frac{d x}{(2)^{2}+(3 x)^{2}}=\frac{1}{3} \int \frac{d t}{(2)^{2}+t^{2}}$

$=\frac{1}{3}\left[\frac{1}{2} \tan ^{-1} \frac{t}{2}\right]$

$=\frac{1}{6} \tan ^{-1}\left(\frac{3 x}{2}\right)$

$=\mathrm{F}(x)$

By second fundamental theorem of calculus, we obtain

$\int_{0}^{2} \frac{d x}{4+9 x^{2}}=\mathrm{F}\left(\frac{2}{3}\right)-\mathrm{F}(0)$

$=\frac{1}{6} \tan ^{-1}\left(\frac{3}{2} \cdot \frac{2}{3}\right)-\frac{1}{6} \tan ^{-1} 0$

$=\frac{1}{6} \tan ^{-1} 1-0$

$=\frac{1}{6} \times \frac{\pi}{4}$

$=\frac{\pi}{24}$

Hence, the correct answer is C.