_ ^ 1 ( e^x + e^ - x ) ^2 ;dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{1}{{{{({e^x} + {e^{ - x}})}^2}}}\;dx = } $

✓
$ - \frac{1}{{2({e^{2x}} + 1)}} + c$
B
$\frac{1}{{2({e^{2x}} + 1)}} + c$
C
$ - \frac{1}{{{e^{2x}} + 1}}$
D
None of these

✓

Answer

Correct option: A.

$ - \frac{1}{{2({e^{2x}} + 1)}} + c$

a
(a)$\int_{}^{} {\frac{1}{{{{({e^x} + {e^{ - x}})}^2}}}\,dx = \int_{}^{} {\frac{{{e^{2x}}}}{{{{({e^{2x}} + 1)}^2}}}} \,dx} $
Put ${e^{2x}} + 1 = t \Rightarrow 2{e^{2x}}dx = dt,$ then it reduces to
$\frac{1}{2}\int_{}^{} {\frac{1}{{{t^2}}}dt} = - \frac{1}{2}.\frac{1}{t} + c = - \frac{1}{{2({e^{2x}} + 1)}} + c.$

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