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

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(A) Let $I = \int \frac{\log \sqrt{x}}{3 x} d x$. Substitute $z = \log \sqrt{x} = \frac{1}{2} \log x$. Then, $dz = \frac{1}{2} \cdot \frac{1}{x} dx$,

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

Vedclass · Answer

(A) Let $I = \int \frac{\log \sqrt{x}}{3 x} d x$. Substitute $z = \log \sqrt{x} = \frac{1}{2} \log x$. Then, $dz = \frac{1}{2} \cdot \frac{1}{x} dx$, which implies $\frac{dx}{x} = 2 dz$. Substituting these into the integral: $I = \int \frac{z}{3} (2 dz) = \frac{2}{3} \int z dz$. Integrating $z$ with respect to $z$: $I = \frac{2}{3} \cdot \frac{z^2}{2} + C = \frac{1}{3} z^2 + C$. Substituting back $z = \log \sqrt{x}$: $I = \frac{1}{3} (\log \sqrt{x})^2 + C$.

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

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