Integral Calculus Question 17
Question: $ \int_{{}}^{{}}{\frac{dx}{e^{x}+{e^{-x}}}=} $
[Bihar CEE 1976; MNR 1974]
Options:
A) $ {{\tan }^{-1}}({e^{-x}}) $
B) $ {{\tan }^{-1}}(e^{x}) $
C) $ \log (e^{x}-{e^{-x}}) $
D) $ \log (e^{x}+{e^{-x}}) $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{dx}{e^{x}+{e^{-x}}}}=\int_{{}}^{{}}{\frac{e^{x}}{e^{2x}+1}},dx=\int_{{}}^{{}}{\frac{dt}{t^{2}+1}}={{\tan }^{-1}}(t) $ $ ={{\tan }^{-1}}(e^{x})+c $ , {Putting $ e^{x}=t\Rightarrow e^{x}dx=dt}. $
BETA
