Question
$\int \frac{\log x}{x} d x$
✓
Answer
Put $\log x=t$
$ \therefore \frac{1}{x} d x= dt$
$\therefore \int \frac{\log x}{x} d x=\int t d t$
$=\frac{ t ^2}{2}+ c$
$\frac{(\log x)^2}{2}+ c $
$ \therefore \frac{1}{x} d x= dt$
$\therefore \int \frac{\log x}{x} d x=\int t d t$
$=\frac{ t ^2}{2}+ c$
$\frac{(\log x)^2}{2}+ c $
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