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

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Solution

$I = \int \frac{log x}{x} d x$

Let $log x = t$

$\frac{1}{x} d x = d t$

Therefore,
$I = \int t d t$

$I = \frac{t^{2}}{2} + C$

$I = \frac{(log x)^{2}}{2} + C$

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