int frac{d x}{e^{x}+e^{-x}} is equal to | vedclass

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Question

(A) Let $I = \int \frac{d x}{e^{x}+e^{-x}}$.Multiply the numerator and denominator by $e^{x}$:$I = \int \frac{e^{x}}{e^{2 x}+1} d x$.Let $e^{x} = t$.

Vedclass · Accepted Answer

$	an ^{-1}\left(e^{x}ight)+C$

Vedclass · Answer

(A) Let $I = \int \frac{d x}{e^{x}+e^{-x}}$.Multiply the numerator and denominator by $e^{x}$:$I = \int \frac{e^{x}}{e^{2 x}+1} d x$.Let $e^{x} = t$. Then $e^{x} d x = d t$.Substituting these into the integral:$I = \int \frac{d t}{t^{2}+1}$.Using the standard integral formula $\int \frac{d x}{x^{2}+1} = 	an ^{-1}(x) + C$:$I = 	an ^{-1}(t) + C$.Substituting $t = e^{x}$ back:$I = 	an ^{-1}\left(e^{x}ight) + C$.Thus,the correct option is $A$.

$\int \frac{d x}{e^{x}+e^{-x}}$ is equal to

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