\(\int \frac{4 e^{x}+6 e^{-x}}{9 e^{x}-4 e^{-x}} d x=A x+B \log \left|9 e^{2 x}-4\right|+c\), then
(Where \(c\) is constant of integration)

\(I=\int \frac{4 e^{2 x}+6}{9 e^{2 x}-4} d x\)
\(4 e^{2 x}+6=A\left(18 e^{2 x}\right)+B\left(9 e^{2 x}-4\right)...(1)\)
\(4 e^{2 x}+6=(18 A+9 B) e^{2 x}-4 B\)
\(\therefore-4 \mathrm{~B}=6 \Rightarrow \mathrm{B}=\frac{-3}{2}\) and \(18 \mathrm{~A}+9 \mathrm{~B}=4\)
\(\therefore \quad 18 \mathrm{~A}-\frac{9 \times 3}{2}=4 \Rightarrow 18 \mathrm{~A}=\frac{35}{2} \Rightarrow \mathrm{A}=\frac{35}{36}\)
\(\therefore 4 e^{2 x}+6=\frac{35}{36}\left(18 e^{2 x}\right)-\frac{3}{2}\left(9 e^{2 x}-4\right)\)
\(I=\int\left[\frac{\frac{35}{36}\left(18 e^{2 x}\right)}{9 e^{2 x}-4}-\frac{\frac{3}{2}\left(9 e^{x}-4\right)}{9 e^{x}-4}\right] d x\)
\(I=\frac{35}{36} \log \left|9 e^{2 x}-4\right|-\frac{3}{2} x+c=A x+B \log \left|9 e^{2 x}-4\right|+c\)
\(\Rightarrow \mathrm{A}=-\frac{3}{2}\) and \(\mathrm{B}=\frac{35}{36}\)