int_a^b frac{log x}{x} , dx = | vedclass

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(C) Let $I = \int_a^b \frac{\log x}{x} \, dx$. Substitute $u = \log x$,then $du = \frac{1}{x} \, dx$. When $x = a$,$u = \log a$. When $x = b$,$u = \lo

Vedclass · Accepted Answer

$\frac{1}{2} \log (ab) \log \left( \frac{b}{a} \right)$

Vedclass · Answer

(C) Let $I = \int_a^b \frac{\log x}{x} \, dx$. Substitute $u = \log x$,then $du = \frac{1}{x} \, dx$. When $x = a$,$u = \log a$. When $x = b$,$u = \log b$. Therefore,$I = \int_{\log a}^{\log b} u \, du$. $I = \left[ \frac{u^2}{2} \right]_{\log a}^{\log b} = \frac{1}{2} [(\log b)^2 - (\log a)^2]$. Using the identity $x^2 - y^2 = (x + y)(x - y)$,we get: $I = \frac{1}{2} [(\log b + \log a)(\log b - \log a)]$. Since $\log b + \log a = \log (ab)$ and $\log b - \log a = \log \left( \frac{b}{a} \right)$,$I = \frac{1}{2} \log (ab) \log \left( \frac{b}{a} \right)$.

$\int_a^b \frac{\log x}{x} \, dx = $

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