int_{0}^{1} frac{dx}{e^{x}+e^{-x}} is equal | vedclass

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Question

(B) To evaluate the integral $ I = \int_{0}^{1} \frac{dx}{e^{x}+e^{-x}} $,we first simplify the integrand by multiplying the numerator and denominator

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$ 	an^{-1}(e)-\frac{\pi}{4} $

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(B) To evaluate the integral $ I = \int_{0}^{1} \frac{dx}{e^{x}+e^{-x}} $,we first simplify the integrand by multiplying the numerator and denominator by $ e^{x} $:$ I = \int_{0}^{1} \frac{e^{x}}{e^{2x}+1} dx $Now,let $ u = e^{x} $. Then $ du = e^{x} dx $.When $ x = 0 $,$ u = e^{0} = 1 $.When $ x = 1 $,$ u = e^{1} = e $.Substituting these into the integral,we get:$ I = \int_{1}^{e} \frac{du}{u^{2}+1} $The integral of $ \frac{1}{u^{2}+1} $ is $ 	an^{-1}(u) $.Evaluating the definite integral:$ I = [	an^{-1}(u)]_{1}^{e} $$ I = 	an^{-1}(e) - 	an^{-1}(1) $Since $ 	an^{-1}(1) = \frac{\pi}{4} $,we have:$ I = 	an^{-1}(e) - \frac{\pi}{4} $

$ \int_{0}^{1} \frac{dx}{e^{x}+e^{-x}} $ is equal to

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