MCQ
$\int_0^1 {\frac{{dx}}{{{e^x} + {e^{ - x}}}}} =$
- A${\tan ^{ - 1}}\left( {\frac{{1 - e}}{{1 + e}}} \right)$
- ✓${\tan ^{ - 1}}\left( {\frac{{e - 1}}{{e + 1}}} \right)$
- C$\frac{\pi }{4}$
- D${\tan ^{ - 1}}e + \frac{\pi }{4}$
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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