MCQ
$\int_{}^{} {\frac{{dx}}{{{e^x} + {e^{ - x}}}} = } $
  • A
    ${\tan ^{ - 1}}({e^{ - x}})$
  • ${\tan ^{ - 1}}({e^x})$
  • C
    $\log ({e^x} - {e^{ - x}})$
  • D
    $\log ({e^x} + {e^{ - x}})$

Answer

Correct option: B.
${\tan ^{ - 1}}({e^x})$
b
(b)$\int_{}^{} {\frac{{dx}}{{{e^x} + {e^{ - x}}}}} = \int_{}^{} {\frac{{{e^x}}}{{{e^{2x}} + 1}}} \,dx = \int_{}^{} {\frac{{dt}}{{{t^2} + 1}}} = {\tan ^{ - 1}}(t)$

$ = {\tan ^{ - 1}}({e^x}) + c$,$\{$Putting ${e^x} = t \Rightarrow {e^x}dx = dt.$$\}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free