_ ^ 1 1 - e^ 2x ;dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{1}{{\sqrt {1 - {e^{2x}}} }}\;dx = } $

✓
$x - \log [1 + \sqrt {1 - {e^{2x}}} ] + c$
B
$x + \log [1 + \sqrt {1 - {e^{2x}}} ] + c$
C
$\log [1 + \sqrt {1 - {e^{2x}}} ] - x + c$
D
None of these

✓

Answer

Correct option: A.

$x - \log [1 + \sqrt {1 - {e^{2x}}} ] + c$

a
(a)$\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 \left[ {t + \sqrt {{t^2} - 1} } \right] + c} $
$ = - \log \left[ {{e^{ - x}} + \sqrt {{e^{ - 2x}} - 1} } \right] = - \log \left[ {\frac{1}{{{e^x}}} + \frac{{\sqrt {1 - {e^{2x}}} }}{{{e^x}}}} \right]$
$ = - \log \left[ {1 + \sqrt {1 - {e^{2x}}} } \right] + \log {e^x} + c$
$ = x - \log \left[ {1 + \sqrt {1 - {e^{2x}}} } \right] + c.$

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