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

MCQ

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

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

✓

Answer

Correct option: A.

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

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

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