_ ^ dx e^x - 1 = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{{e^x} - 1}} = } $

✓
$\ln (1 - {e^{ - x}}) + c$
B
$ - \ln (1 - {e^{ - x}}) + c$
C
$\ln ({e^x} - 1) + c$
D
None of these

✓

Answer

Correct option: A.

$\ln (1 - {e^{ - x}}) + c$

a
(a)$\int_{}^{} {\frac{{dx}}{{{e^x} - 1}} = \int_{}^{} {\frac{{{e^{ - x}}}}{{1 - {e^{ - x}}}}\,dx} } $
Put $1 - {e^{ - x}} = t \Rightarrow {e^{ - x}}dx = dt,$ then it reduces to
$\int_{}^{} {\frac{{dt}}{t} = \log t + c} = \log (1 - {e^{ - x}}) + c.$

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