int frac{log sqrt{x}}{3 x} dx is equal to | vedclass

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(A) Let $I = \int \frac{\log \sqrt{x}}{3 x} dx$. Using the property $\log \sqrt{x} = \log(x^{1/2}) = \frac{1}{2} \log x$, the integral becomes: $I = \

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$\frac{1}{3}(\log \sqrt{x})^2+C$

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(A) Let $I = \int \frac{\log \sqrt{x}}{3 x} dx$. Using the property $\log \sqrt{x} = \log(x^{1/2}) = \frac{1}{2} \log x$, the integral becomes: $I = \int \frac{\frac{1}{2} \log x}{3 x} dx = \frac{1}{6} \int \frac{\log x}{x} dx$. Let $u = \log x$, then $du = \frac{1}{x} dx$. Substituting these into the integral: $I = \frac{1}{6} \int u du = \frac{1}{6} \cdot \frac{u^2}{2} + C = \frac{u^2}{12} + C$. Substituting $u = \log x$ back: $I = \frac{(\log x)^2}{12} + C$. Alternatively, using the substitution $t = \log \sqrt{x} = \frac{1}{2} \log x$, then $dt = \frac{1}{2x} dx$, which implies $\frac{1}{x} dx = 2 dt$. $I = \int \frac{t}{3} (2 dt) = \frac{2}{3} \int t dt = \frac{2}{3} \cdot \frac{t^2}{2} + C = \frac{t^2}{3} + C = \frac{1}{3} (\log \sqrt{x})^2 + C$. Thus, the correct option is $A$.

$\int \frac{\log \sqrt{x}}{3 x} dx$ is equal to

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