Question: $\frac{1}{x(\log x)^{m}}, x>0$
Solution:
Let $\log x=t$
$\therefore \frac{1}{x} d x=d t$
$\Rightarrow \int \frac{1}{x(\log x)^{m}} d x=\int \frac{d t}{(t)^{m}}$\
$=\left(\frac{t^{-m+1}}{1-m}\right)+\mathrm{C}$
$=\frac{(\log x)^{1-m}}{(1-m)}+\mathrm{C}$
