Integral Calculus Question 233
Question: $ \int_{{}}^{{}}{\frac{dx}{e^{x}-1}=} $
[MP PET 1989]
Options:
A) $ \ln (1-{e^{-x}})+c $
B) $ -\ln (1-{e^{-x}})+c $
C) $ \ln (e^{x}-1)+c $
D) None of these
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Answer:
Correct Answer: A
Solution:
$ \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. $
BETA
