If $\int\frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}dx=Ax+B\log(9e^{2x}-4)+C$, then:
Answer & explanation
Correct answer: option 3
$\int\frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}dx=\int\frac{4e^{2x}+6}{9e^{2x}-4}dx=\int\frac{4e^{2x}}{9e^{2x}-4}dx+6\int\frac{1}{9e^{2x}-4}dx$
$=\frac{2}{9}\int\frac{18e^{2x}}{9e^{2x}-4}dx+6\int\frac{e^{-2x}}{9-4e^{-2x}}dx=\frac{2}{9}\log(9e^{2x}-4)+\frac{6}{8}\log(9-4e^{-2x})+C$
$=\frac{2}{9}\log(9e^{2x}-4)+\frac{3}{4}\log e^{2x}+C=\frac{-3}{2}x+\frac{35}{36}\log(9e^{2x}-4)+C$