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$\int \frac{1}{(1 + e^{x}) (1 + e^{- x})} d x =$

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$\frac{- 1}{1 + e^{x}} + c$

$\frac{1}{1 + e^{x}} + c$

$\frac{1}{1 - e^{x}} + c$

$\frac{- 2}{1 + e^{x}} + c$

answer is A.

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Detailed Solution

$\int \frac{1}{(1 + e^{x}) (1 + e^{- x})} d x$

$= \int \frac{1}{(1 + e^{x}) \frac{(e^{x} + 1)}{e^{x}}} d x = \int \frac{e^{x}}{{(1 + e^{x})}^{2}} d x$

$= \int \frac{d t}{{(1 + t)}^{2}}$ put $e^{x} = t \Rightarrow e^{x} d x = d t$

$= \frac{- 1}{1 + t} + c$

$= \frac{- 1}{1 + e^{x}} + c$

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