The value of _0^1 dx e^x + e^ - x is - Vidyadip

MCQ

The value of $\int_0^1 {\frac{{dx}}{{{e^x} + {e^{ - x}}}}} $ is

A
${\tan ^{ - 1}}\left( {\frac{{1 - e}}{{1 + e}}} \right)$
✓
${\tan ^{ - 1}}\left( {\frac{{e - 1}}{{e + 1}}} \right)$
C
$\frac{\pi }{4}$
D
${\tan ^{ - 1}}e + \frac{\pi }{4}$

✓

Answer

Correct option: B.

${\tan ^{ - 1}}\left( {\frac{{e - 1}}{{e + 1}}} \right)$

b
(b) $\int_0^1 {\frac{{dx}}{{{e^x} + {e^{ - x}}}} = \int_0^1 {\frac{{{e^x}}}{{1 + {e^{2x}}}}dx} } $

Now put ${e^x} = t \Rightarrow {e^x}dx = dt$

Also as $x = 0$ to $1$, $t = 1$ to $e$, then reduced form is

$\int_1^e {\frac{{dt}}{{1 + {t^2}}} = [{{\tan }^{ - 1}}t]_1^e} = {\tan ^{ - 1}}\left( {\frac{{e - 1}}{{e + 1}}} \right)$,

$\left[ \because {{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{x-y}{1+xy} \right) \right]$.

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