Integrate the function $\frac{x}{9-4x^{2}}$.

(N/A) Let $I = \int \frac{x}{9-4x^{2}} dx$.
Substitute $t = 9-4x^{2}$.
Then,$dt = -8x dx$,which implies $x dx = -\frac{1}{8} dt$.
Substituting these into the integral,we get:
$I = \int \frac{1}{t} \left(-\frac{1}{8}\right) dt$
$I = -\frac{1}{8} \int \frac{1}{t} dt$
$I = -\frac{1}{8} \log |t| + C$
Substituting back $t = 9-4x^{2}$,we get:
$I = -\frac{1}{8} \log |9-4x^{2}| + C$,where $C$ is an arbitrary constant.