$\cfrac{1}{{x{{\left( {\log x} \right)}^m}}},x > 0$

.: Let $I = \int {\cfrac{1}{{x{{\left( {\log x} \right)}^m}}}} dx$

Put $\log x = t$ $\Rightarrow$ $\cfrac{1}{x}dx = dt$

$\therefore$ $I = \int {\cfrac{{dt}}{{{t^m}}}} = \int {{t^{ - m}}dt} = \cfrac{{{t^{ - m + 1}}}}{{ - m + 1}} + C = \cfrac{{{{\left( {{\mathop{\rm logx}\nolimits} } \right)}^{1 - m}}}}{{1 - m}} + C$