int_{frac{2}{e}}^{frac{1}{e}} frac{1}{x(log x | vedclass

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Question

(D) Let $I = \int_{\frac{2}{e}}^{\frac{1}{e}} \frac{1}{x(\log x)^{1/3}} dx$.Substitute $t = \log x$,then $dt = \frac{1}{x} dx$.When $x = \frac{1}{e}$,

Vedclass · Accepted Answer

$\frac{3}{2}\left\{1-(\log 2-1)^{\frac{2}{3}}\right\}$

Vedclass · Answer

(D) Let $I = \int_{\frac{2}{e}}^{\frac{1}{e}} \frac{1}{x(\log x)^{1/3}} dx$.Substitute $t = \log x$,then $dt = \frac{1}{x} dx$.When $x = \frac{1}{e}$,$t = \log(\frac{1}{e}) = -1$.When $x = \frac{2}{e}$,$t = \log(\frac{2}{e}) = \log 2 - \log e = \log 2 - 1$.Thus,$I = \int_{\log 2 - 1}^{-1} t^{-1/3} dt$.Integrating,we get $I = \left[ \frac{t^{2/3}}{2/3} \right]_{\log 2 - 1}^{-1} = \frac{3}{2} \left[ t^{2/3} \right]_{\log 2 - 1}^{-1}$.$I = \frac{3}{2} \left[ (-1)^{2/3} - (\log 2 - 1)^{2/3} \right]$.Since $(-1)^{2/3} = ((-1)^2)^{1/3} = 1^{1/3} = 1$,we have $I = \frac{3}{2} \left[ 1 - (\log 2 - 1)^{2/3} \right]$.

$\int_{\frac{2}{e}}^{\frac{1}{e}} \frac{1}{x(\log x)^{\frac{1}{3}}} dx$ is equal to

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