(\integral\_{}^{}{\frac{dx}{e^{x} + e^{- x}} =}) | MATH - INTEGRATION BY SUBSTITUTION | SolveeIt

$\int_{}^{}{\frac{dx}{e^{x} + e^{- x}} =}$

$\tan^{- 1}(e^{- x})$

$\tan^{- 1}(e^{x})$

$\log(e^{x} - e^{- x})$

None of these

Answer

$\tan^{- 1}(e^{- x})$

Explanation

Put $2a^{\sqrt{x}}\log_{10}a + c$

$2a^{\sqrt{x}}\log_{a}10 + c$ then it reduces to

$\int_{}^{}{\frac{x^{3}}{\sqrt{1 - x^{8}}}dx =}$

Question: \(\int_{}^{}{\frac{dx}{e^{x} + e^{- x}} =}\)...