Integral Calculus Question 414
Question: $ \int_{{}}^{{}}{\frac{1}{\sqrt{1-e^{2x}}}\ dx=} $
[MP PET 1993, 2002; RPET 1999]
Options:
A) $ x-\log [1+\sqrt{1-e^{2x}}]+c $
B) $ x+\log [1+\sqrt{1-e^{2x}}]+c $
C) $ \log [1+\sqrt{1-e^{2x}}]-x+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{1}{\sqrt{1-e^{2x}}},dx=\int_{{}}^{{}}{\frac{{e^{-x}}}{\sqrt{{e^{-2x}}-1}},dx}} $ Put $ {e^{-x}}=t\Rightarrow -{e^{-x}}dx=dt, $ then it reduces to $ -\int_{{}}^{{}}{\frac{1}{\sqrt{t^{2}-1}},dt=-\log [ t+\sqrt{t^{2}-1} ]+c} $ $ =-\log [ {e^{-x}}+\sqrt{{e^{-2x}}-1} ]=-\log [ \frac{1}{e^{x}}+\frac{\sqrt{1-e^{2x}}}{e^{x}} ] $ $ =-\log [ 1+\sqrt{1-e^{2x}} ]+\log e^{x}+c $ $ =x-\log [ 1+\sqrt{1-e^{2x}} ]+c. $
BETA
