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$\int_2^3 \frac{\log x}{x} d x=$
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The correct answer is:
$\frac{1}{2} \log 6 \log \frac{3}{2}$
$\begin{aligned} & \int_2^3 \frac{\log x}{x} d x=\left[\frac{(\log x)^2}{2}\right]_2^3 \\ & =\frac{1}{2}\left\{(\log 3)^2-(\log 2)^2\right\} \\ & =\frac{1}{2}\{\log 3+\log 2\}\{\log 3-\log 2\} \\ & =\frac{1}{2} \log 6 \cdot \log \frac{3}{2}\end{aligned}$
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