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$\int \frac{d x}{1+e^x}=$
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Verified Answer
The correct answer is:
$-\log \left(1+e^{-x}\right)$
$\int \frac{d x}{1+e^x}=\int \frac{e^{-x}}{1+e^{-x}} dx$
Put $1+e^{-x}=t \Rightarrow e^{-x} d x=-d t$, then it reduces to $-\int \frac{d t}{t}=-\log t=-\log \left(1+e^{-x}\right)$
Put $1+e^{-x}=t \Rightarrow e^{-x} d x=-d t$, then it reduces to $-\int \frac{d t}{t}=-\log t=-\log \left(1+e^{-x}\right)$
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