Integral Calculus Question 4
Question: $ \int_{{}}^{{}}{\frac{dx}{1+e^{x}}=} $
[MP PET 1991; Roorkee 1977]
Options:
A) $ \log (1+e^{x}) + c$
B) $ -\log (1+{e^{-x}}) + c$
C) $ -\log (1-{e^{-x}}) + c$
D) $ \log ({e^{-x}}+{e^{-2x}}) + c$
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{dx}{1+e^{x}}}=\int_{{}}^{{}}{\frac{{e^{-x}}}{1+{e^{-x}}}},dx $
Put $ 1+{e^{-x}}=t $
$ {e^{-x}}dx=-dt $ , then it reduces to
$ -\int{\frac{dt}{t}=-\log t=-\log (1+{e^{-x}})} + c$ .
BETA
