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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.)
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$\frac{3}{4}, \frac{-1}{24}$
$\begin{aligned} & \int \frac{2 e^x+3 e^{-x}}{3 e^x+4 e^{-x}} d x=\int \frac{2 e^{2 x}+3}{3 e^{2 x}+4} d x \\ & \int \frac{\frac{3}{4}\left(3 e^{2 x}+4\right)-\frac{1}{24}\left(6 e^{2 x}\right)}{3 e^{2 x}+4} d x \\ & =\frac{3}{4} x-\frac{1}{24} \log \left(3 e^{2 x}+4\right)+c \\ & \Rightarrow A=\frac{3}{4} \text { and } B=\frac{-1}{24}\end{aligned}$
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