2ex−3=t 2exdx=dt 12∫1t2dt=12−1t+C =−121(2ex−3)+C

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

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

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

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

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

answer is B.

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Detailed Solution

$2 e^{x} - 3 = t$

$2 e^{x} d x = d t$

$\frac{1}{2} \int \frac{1}{t^{2}} d t = \frac{1}{2} \frac{- 1}{t} + C$

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

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