MCQ
$\int_{}^{} {\frac{{dx}}{{{e^x} + {e^{ - x}}}} = } $
- A${\tan ^{ - 1}}({e^{ - x}})$
- ✓${\tan ^{ - 1}}({e^x})$
- C$\log ({e^x} - {e^{ - x}})$
- D$\log ({e^x} + {e^{ - x}})$
$ = {\tan ^{ - 1}}({e^x}) + c$,$\{$Putting ${e^x} = t \Rightarrow {e^x}dx = dt.$$\}$
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