Definite Integration Question 361
Question: $ \int_0^{1}{\frac{{e^{-x}}}{1+{e^{-x}}}}dx= $
[Roorkee 1976]
Options:
A) $ \log ( \frac{1+e}{e} )-\frac{1}{e}+1 $
B) $ \log ( \frac{1+e}{2e} )-\frac{1}{e}+1 $
C) $ \log ( \frac{1+e}{2e} )+\frac{1}{e}-1 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Put $ 1+{e^{-x}}=t\Rightarrow -{e^{-x}}dx=dt $ , then we have $ I=\int_2^{1+\frac{1}{e}}{\frac{(t-1)(-dt)}{t}}=\int_2^{1+\frac{1}{e}}{( \frac{1}{t}-1 )}dt $
$ =[ {\log _{e}}t-t ]_2^{1+\frac{1}{e}}={\log _{e}}( 1+\frac{1}{e} )-( 1+\frac{1}{e} )-{\log _{e}}2+2 $
$ ={\log _{e}}( \frac{e+1}{2e} )-\frac{1}{e}+1 $ .
BETA
