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

MCQ

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

✓
$\frac{{ - 1}}{{2({e^{2x}} + 1)}} + c$
B
$\frac{1}{{2({e^{2x}} + 1)}} + c$
C
$\frac{1}{{{e^{2x}} + 1}} + c$
D
$\frac{{ - 1}}{{{e^{2x}} + 1}} + c$

✓

Answer

Correct option: A.

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

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

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