$\int {\cfrac{{dx}}{{{e^x} + {e^{ - x}}}}}$is equal to

Option a is correct

: Let $I = \int {\cfrac{{dx}}{{{e^x} + {e^{ - x}}}} = } \int {\cfrac{{dx}}{{{e^x} + \cfrac{1}{{{e^x}}}}}}$

$\Rightarrow$ $\int {\cfrac{{{e^x}dx}}{{{{\left( {{e^x}} \right)}^2} + 1}}}$

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

$\therefore$ $I = \int {\cfrac{{dt}}{{1 + {t^2}}}} = {\tan ^{ - 1}}\left( t \right) + C = {\tan ^{ - 1}}\left( {{e^x}} \right) + C$