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$\int_0^1 \frac{e^{-x}}{1+e^{-x}} d x=$
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Verified Answer
The correct answer is:
$\log \left(\frac{1+e}{2 e}\right)-\frac{1}{e}+1$
Put $1+e^{-x}=t \Rightarrow-e^{-x} d x=d t$, then we have
$\begin{aligned} I & =\int_2^{1+\frac{1}{e}} \frac{(t-1)(-d t)}{t}=\int_2^{1+\frac{1}{e}}\left(\frac{1}{t}-1\right) d t \\ & =\left[\log _e t-t\right]_2^{1+\frac{1}{e}}=\log _e\left(1+\frac{1}{e}\right)-\left(1+\frac{1}{e}\right)-\log _e 2+2 \\ & =\log _\theta\left(\frac{e+1}{2 e}\right)-\frac{1}{e}+1\end{aligned}$
$\begin{aligned} I & =\int_2^{1+\frac{1}{e}} \frac{(t-1)(-d t)}{t}=\int_2^{1+\frac{1}{e}}\left(\frac{1}{t}-1\right) d t \\ & =\left[\log _e t-t\right]_2^{1+\frac{1}{e}}=\log _e\left(1+\frac{1}{e}\right)-\left(1+\frac{1}{e}\right)-\log _e 2+2 \\ & =\log _\theta\left(\frac{e+1}{2 e}\right)-\frac{1}{e}+1\end{aligned}$
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