Integral Calculus Question 485
Question: The value of $ \int_{{}}^{{}}{\frac{e^{x}}{e^{x}+1}},dx $ is
[Pb. CET 2000]
Options:
A) $ e^{x}+c $
B) $ (e^{x}+1)+c $
C) $ \log (e^{x}+1)+c $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Put $ e^{x}+1=t\Rightarrow e^{x}dx=dt $
$ \therefore $ $ \int_{{}}^{{}}{\frac{e^{x}}{e^{x}+1}dx=\int_{{}}^{{}}{\frac{dt}{t}=\log t+c=\log (e^{x}+1)+c}} $ .
BETA
