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$\int \frac{\log \sqrt{x}}{3 x} d x$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{3}(\log \sqrt{x})^{2}+C$
Let $I=\int \frac{\log \sqrt{x}}{3 x} d x$
Aqain, let $\log \sqrt{x}=z \Rightarrow \frac{1}{2 x} d x=d z$
$I=\int \frac{2 z}{3} d z=\frac{2}{3} \int z d z$
$=\frac{2}{3} \cdot \frac{z^{2}}{2}+C=\frac{1}{3} \left(\log \sqrt{x}\right)^{2}+C$
Aqain, let $\log \sqrt{x}=z \Rightarrow \frac{1}{2 x} d x=d z$
$I=\int \frac{2 z}{3} d z=\frac{2}{3} \int z d z$
$=\frac{2}{3} \cdot \frac{z^{2}}{2}+C=\frac{1}{3} \left(\log \sqrt{x}\right)^{2}+C$
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