Prove

Let $9-4 x^{2}=t$

$\therefore-8 x d x=d t$

$\Rightarrow \int \frac{x}{9-4 x^{2}} d x=\frac{-1}{8} \int \frac{1}{t} d t$

$=\frac{-1}{8} \log |t|+C$

$=\frac{-1}{8} \log \left|9-4 x^{2}\right|+C$