Integral Calculus Question 467
Question: What is the value of the integral $ I=\int{\frac{dx}{(1+e^{x})(1+{e^{-x}})}} $
[DCE 1999]
Options:
A) $ \frac{-1}{1+e^{x}} $
B) $ \frac{e^{x}}{1+e^{x}} $
C) $ \frac{1}{1+e^{x}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int{\frac{dx}{(1+e^{x})( 1+\frac{1}{e^{x}} )}=\int{\frac{e^{x}dx}{{{(1+e^{x})}^{2}}}}} $ Let $ 1+e^{x}=t $ ,
$ \therefore e^{x},dx=dt $ \ $ I=\int{\frac{dt}{t^{2}}=\int{{t^{-2}}dt=\frac{{t^{-1}}}{-1}=\frac{{{(1+e^{x})}^{-1}}}{-1}}}=\frac{-1}{1+e^{x}} $ .
BETA
