int_{1}^{3} left(frac{x^{2}+1}{4x}right)^{-1} | vedclass

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Question

(C) We are given the integral $ I = \int_{1}^{3} \left(\frac{x^{2}+1}{4x}\right)^{-1} dx $. Simplifying the integrand,we get $ I = \int_{1}^{3} \frac{

Vedclass · Accepted Answer

$ \log 25 $

Vedclass · Answer

(C) We are given the integral $ I = \int_{1}^{3} \left(\frac{x^{2}+1}{4x}\right)^{-1} dx $. Simplifying the integrand,we get $ I = \int_{1}^{3} \frac{4x}{x^{2}+1} dx $. Let $ u = x^{2}+1 $,then $ du = 2x dx $,which implies $ 2 du = 4x dx $. When $ x = 1 $,$ u = 1^{2}+1 = 2 $. When $ x = 3 $,$ u = 3^{2}+1 = 10 $. Substituting these into the integral,we get $ I = \int_{2}^{10} \frac{2}{u} du $. $ I = 2 [\ln |u|]_{2}^{10} $. $ I = 2 (\ln 10 - \ln 2) $. $ I = 2 \ln \left(\frac{10}{2}\right) = 2 \ln 5 $. Using the property $ n \ln a = \ln(a^{n}) $,we get $ I = \ln(5^{2}) = \ln 25 $. Thus,the correct option is $ C $.

$\int_{1}^{3} \left(\frac{x^{2}+1}{4x}\right)^{-1} dx = $ . . . . . . .

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