int_{1}^{e} frac{dx}{x(1+log x)^{2}} = | vedclass

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Question

(A) Let $I = \int_{1}^{e} \frac{dx}{x(1+\log x)^{2}}$.Substitute $u = 1 + \log x$. Then $du = \frac{1}{x} dx$.When $x = 1$,$u = 1 + \log 1 = 1 + 0 = 1

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$\frac{1}{2}$

Vedclass · Answer

(A) Let $I = \int_{1}^{e} \frac{dx}{x(1+\log x)^{2}}$.Substitute $u = 1 + \log x$. Then $du = \frac{1}{x} dx$.When $x = 1$,$u = 1 + \log 1 = 1 + 0 = 1$.When $x = e$,$u = 1 + \log e = 1 + 1 = 2$.Thus,$I = \int_{1}^{2} \frac{du}{u^{2}} = \int_{1}^{2} u^{-2} du$.$I = \left[ \frac{u^{-1}}{-1} \right]_{1}^{2} = \left[ -\frac{1}{u} \right]_{1}^{2}$.$I = -\left( \frac{1}{2} - \frac{1}{1} \right) = -\left( -\frac{1}{2} \right) = \frac{1}{2}$.

$\int_{1}^{e} \frac{dx}{x(1+\log x)^{2}} =$

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