_0^1 e^ - x 1 + e^ - x ,dx = - Vidyadip

MCQ

$\int_0^1 {\frac{{{e^{ - x}}}}{{1 + {e^{ - x}}}}} \,dx = $

A
$\log \left( {\frac{{1 + e}}{e}} \right) - \frac{1}{e} + 1$
✓
$\log \left( {\frac{{1 + e}}{{2e}}} \right) - \frac{1}{e} + 1$
C
$\log \left( {\frac{{1 + e}}{{2e}}} \right) + \frac{1}{e} - 1$
D
None of these

✓

Answer

Correct option: B.

$\log \left( {\frac{{1 + e}}{{2e}}} \right) - \frac{1}{e} + 1$

b
(b) Put $1 + {e^{ - x}} = t $

$\Rightarrow - {e^{ - x}}dx = dt$, then we have

$I = \int_2^{1 + \frac{1}{e}} {\frac{{(t - 1)( - dt)}}{t}} $

$= \int_2^{1 + \frac{1}{e}} {\left( {\frac{1}{t} - 1} \right)} \,dt$

$ = \left[ {{{\log }_e}t - t} \right]_2^{1 + \frac{1}{e}} $

$= {\log _e}\left( {1 + \frac{1}{e}} \right) - \left( {1 + \frac{1}{e}} \right) - {\log _e}2 + 2$

$ = {\log _e}\left( {\frac{{e + 1}}{{2e}}} \right) - \frac{1}{e} + 1$.

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