If $\int \frac{2 e^x+3 e^{-x}}{3 e^x+4 e^{-x}} d x=A x+B \log \left(3 e^{2 x}+4\right)+C$,then values of $A$ and $B$ are respectively (where $C$ is a constant of integration.)

  • A
    $\frac{3}{4}, \frac{-1}{24}$
  • B
    $\frac{3}{4}, \frac{1}{24}$
  • C
    $\frac{4}{3}, -24$
  • D
    $\frac{1}{4}, \frac{1}{24}$

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