int_2^3 frac{log x}{x} d x= | vedclass

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Question

(C) Let $I = \int_2^3 \frac{\log x}{x} d x$. Substitute $u = \log x$,then $du = \frac{1}{x} dx$. When $x = 2$,$u = \log 2$. When $x = 3$,$u = \log 3$.

Vedclass · Accepted Answer

$\frac{1}{2} \log 6 \log \frac{3}{2}$

Vedclass · Answer

(C) Let $I = \int_2^3 \frac{\log x}{x} d x$. Substitute $u = \log x$,then $du = \frac{1}{x} dx$. When $x = 2$,$u = \log 2$. When $x = 3$,$u = \log 3$. Thus,$I = \int_{\log 2}^{\log 3} u du = \left[ \frac{u^2}{2} ight]_{\log 2}^{\log 3}$. $I = \frac{1}{2} \left( (\log 3)^2 - (\log 2)^2 ight)$. Using the identity $a^2 - b^2 = (a+b)(a-b)$,we get: $I = \frac{1}{2} (\log 3 + \log 2)(\log 3 - \log 2)$. Since $\log a + \log b = \log(ab)$ and $\log a - \log b = \log(\frac{a}{b})$,$I = \frac{1}{2} \log(3 	imes 2) \log(\frac{3}{2}) = \frac{1}{2} \log 6 \log \frac{3}{2}$.

$\int_2^3 \frac{\log x}{x} d x=$

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