The integral of the function $\frac{1}{9-4x^2}$ is:

$\frac{1}{12}\log_e\left|\frac{3+2x}{3-2x}\right|+C$, where C is an arbitrary constant

The correct answer is Option (2) → $\frac{1}{12}\log_e\left|\frac{3+2x}{3-2x}\right|+C$, where C is an arbitrary constant

$\int\frac{1}{9-4x^2}dx=\frac{1}{4}\int\frac{1}{(\frac{3}{2})^2-x^2}dx$

$=\frac{1}{4}×\frac{2}{2×3}\log\left|\frac{\frac{3}{2}+x}{\frac{3}{2}-x}\right|+C$

$=\frac{1}{12}\log\left|\frac{3+2x}{3-2x}\right|+C$