Integral Calculus Question 413
Question: $ \int_{{}}^{{}}{\frac{{e^{-x}}}{1+e^{x}}\ dx=} $
Options:
A) $ \log (1+e^{x})-x-{e^{-x}}+c $
B) $ \log (1+e^{x})+x-{e^{-x}}+c $
C) $ \log (1+e^{x})-x+{e^{-x}}+c $
D) $ \log (1+e^{x})+x+{e^{-x}}+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{{e^{-x}}}{1+e^{x}},dx}=\int_{{}}^{{}}{\frac{{e^{-x}}{e^{-x}}}{{e^{-x}}+1},dx} $ Put $ {e^{-x}}+1=t\Rightarrow -{e^{-x}}dx=dt, $ then it reduces to $ -\int_{{}}^{{}}{\frac{(t-1)}{t},dt}=\int_{{}}^{{}}{( \frac{1}{t}-1 ),dt} $ $ =\log t-t+c=\log ({e^{-x}}+1)-({e^{-x}}+1)+c $ $ =\log (e^{x}+1)-x-{e^{-x}}-1+c $ $ =\log (e^{x}+1)-x-{e^{-x}}+c $ , $ (\because ,1= $ constant).
BETA
