The value of int_0^1 frac{dx}{e^x + e^{-x}} i | vedclass

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(B) We have the integral $I = \int_0^1 \frac{dx}{e^x + e^{-x}}$.Multiply the numerator and denominator by $e^x$:$I = \int_0^1 \frac{e^x}{e^{2x} + 1} d

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

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(B) We have the integral $I = \int_0^1 \frac{dx}{e^x + e^{-x}}$.Multiply the numerator and denominator by $e^x$:$I = \int_0^1 \frac{e^x}{e^{2x} + 1} dx$.Let $t = e^x$,then $dt = e^x dx$.When $x = 0$,$t = e^0 = 1$. When $x = 1$,$t = e^1 = e$.Substituting these into the integral:$I = \int_1^e \frac{dt}{1 + t^2} = [	an^{-1} t]_1^e$.Evaluating the limits:$I = 	an^{-1}(e) - 	an^{-1}(1) = 	an^{-1}(e) - \frac{\pi}{4}$.Using the identity $	an^{-1} x - 	an^{-1} y = 	an^{-1}\left(\frac{x - y}{1 + xy}ight)$:$I = 	an^{-1}\left(\frac{e - 1}{1 + e \cdot 1}ight) = 	an^{-1}\left(\frac{e - 1}{e + 1}ight)$.

The value of $\int_0^1 \frac{dx}{e^x + e^{-x}}$ is

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