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

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Question

(A) Given integral: $I = \int_{1}^{x} \frac{\log(x^2)}{x} \, dx$ Using the property $\log(a^b) = b \log a$,we have: $I = \int_{1}^{x} \frac{2 \log x}{

Vedclass · Accepted Answer

$(\log x)^2$

Vedclass · Answer

(A) Given integral: $I = \int_{1}^{x} \frac{\log(x^2)}{x} \, dx$ Using the property $\log(a^b) = b \log a$,we have: $I = \int_{1}^{x} \frac{2 \log x}{x} \, dx$ Let $t = \log x$,then $dt = \frac{1}{x} \, dx$. When $x = 1$,$t = \log(1) = 0$. When $x = x$,$t = \log x$. Substituting these into the integral: $I = \int_{0}^{\log x} 2t \, dt$ $I = 2 \left[ \frac{t^2}{2} \right]_{0}^{\log x}$ $I = [t^2]_{0}^{\log x}$ $I = (\log x)^2 - 0^2 = (\log x)^2$ Thus,the correct option is $A$.

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

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