If \(2x>3\), then \(\int \frac{dx}{9-4x^2}=\)

\(\frac{1}{12}\log_{e} \frac{3-2x}{3+2x}+c\) \(\frac{1}{12}\log_{e} \frac{3+2x}{3-2x}+c\) \(\frac{1}{12}\log_{e} \frac{2x+3}{2x-3}+c\) \(\frac{1}{12}\log_{e} \frac{2x-3}{2x+3}+c\)

\(\frac{1}{12}\log_{e} \frac{2x+3}{2x-3}+c\)

\(\begin{aligned}\int \frac{dx}{9-4x^2}&=\int \frac{dx}{3^2-(2x)^2}\\ &=\frac{1}{4}\int \frac{dx}{\left(\frac{3}{2}\right)^2-x^2}\\ &=\frac{1}{4}\cdot \frac{1}{3}\log \left(\frac{3+2x}{3-2x}\right)+c\end{aligned}\)