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

log | e^{x} - 1 | - log | e^{x} + 2 | + c

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\frac{1}{2} log | e^{x} - 1 | - \frac{1}{3} log | e^{x} + 2 | + c

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\frac{1}{3} log | e^{x} - 1 | - \frac{1}{3} log | e^{x} + 2 | + c

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\frac{1}{3} log | e^{x} - 1 | + \frac{1}{3} log | e^{x} + 2 | + c

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Solution

The correct option is C $\frac{1}{3} log | e^{x} - 1 | - \frac{1}{3} log | e^{x} + 2 | + c$
Consider, $I = \int \frac{1}{e^{x} + 1 - 2 e^{- x}} = \int \frac{e^{x}}{e^{2 x} + e^{x} - 2} d x$

$(e^{x} - 1) (e^{x} + 2) = e^{2 x} + e^{x} - 2$

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

$= \int \frac{e^{x}}{(e^{x} - 1) (e^{x} + 2)} d x = \int \frac{e^{x}}{3} [\frac{1}{e^{x} - 1} - \frac{1}{e^{x} + 2}] d x$

$= \frac{1}{3} \int \frac{e^{x}}{(e^{x} - 1)} d x - \frac{1}{3} \int \frac{e^{x}}{(e^{x} + 2)} d x$

Let $e^{x} - 1 = t$

$e^{x} + 2 = u$

$e^{x} d x = d t$

$e^{x} d x = d u$

$= \frac{1}{3} log (t) - \frac{1}{3} log u + c$

$= \frac{1}{3} log (e^{x} - 1) - \frac{1}{3} log (e^{x} + 2) + C$

$= \frac{1}{3} [log | e^{x} - 1 | - log | e^{x} + 2 |] + C$
Option C

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