Evaluate the integral: int frac{dx}{e^x + e^{ | vedclass

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(D) To evaluate the integral $I = \int \frac{dx}{e^x + e^{-x}}$,we first simplify the integrand: $I = \int \frac{dx}{e^x + \frac{1}{e^x}} = \int \frac

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$	an^{-1}(e^x)$

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(D) To evaluate the integral $I = \int \frac{dx}{e^x + e^{-x}}$,we first simplify the integrand: $I = \int \frac{dx}{e^x + \frac{1}{e^x}} = \int \frac{e^x dx}{(e^x)^2 + 1}$. Let $u = e^x$. Then $du = e^x dx$. Substituting these into the integral,we get: $I = \int \frac{du}{u^2 + 1}$. Using the standard integral formula $\int \frac{du}{u^2 + 1} = 	an^{-1}(u) + C$,we obtain: $I = 	an^{-1}(e^x) + C$. Therefore,the correct option is $D$.

Evaluate the integral: $\int \frac{dx}{e^x + e^{-x}} = $ . . . . . . $+ C$.

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