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$\int \frac{1+2 e^{-x}}{1-2 e^{-x}} d x=$
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Verified Answer
The correct answer is:
$x+2\log \left(1-2 e^{-x}\right)+\mathrm{c}$
(B)
Let $\begin{aligned} 1 &=\int \frac{1-2 e^{-x}}{1-2 e^{-1}} \\=& \int \frac{1-2 e^{-x}+4 e^{-x}}{1-2 e^{-1}} d x=\int \frac{1-2 e^{-x}}{1-2 e^{-1}} d x+4 \int \frac{e^{-x}}{1-2 e^{-x}} d x \\=& \int d x+\frac{4}{2} \int \frac{2 e^{-x}}{1-2 e^{-x}} d x=x-2 \cdot \log 1-2 e^{-1}-c \end{aligned}$
Let $\begin{aligned} 1 &=\int \frac{1-2 e^{-x}}{1-2 e^{-1}} \\=& \int \frac{1-2 e^{-x}+4 e^{-x}}{1-2 e^{-1}} d x=\int \frac{1-2 e^{-x}}{1-2 e^{-1}} d x+4 \int \frac{e^{-x}}{1-2 e^{-x}} d x \\=& \int d x+\frac{4}{2} \int \frac{2 e^{-x}}{1-2 e^{-x}} d x=x-2 \cdot \log 1-2 e^{-1}-c \end{aligned}$
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