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

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(D) Let $I = \int \frac{\log \sqrt{x}}{3 x} \,d x$.We know that $\log \sqrt{x} = \log (x^{1/2}) = \frac{1}{2} \log x$.Substituting this into the integ

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$\frac{1}{12}(\log x)^2+c$,(where $c$ is a constant of integration)

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(D) Let $I = \int \frac{\log \sqrt{x}}{3 x} \,d x$.We know that $\log \sqrt{x} = \log (x^{1/2}) = \frac{1}{2} \log x$.Substituting this into the integral:$I = \int \frac{\frac{1}{2} \log x}{3 x} \,d x = \frac{1}{6} \int \frac{\log x}{x} \,d x$.Let $u = \log x$,then $du = \frac{1}{x} \,d x$.The integral becomes $I = \frac{1}{6} \int u \,du = \frac{1}{6} \cdot \frac{u^2}{2} + c = \frac{u^2}{12} + c$.Substituting $u = \log x$ back,we get $I = \frac{1}{12} (\log x)^2 + c$.

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

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