Question

integrate the function [1/x(logx)^m].dx, x > 0

Answer 1

given, $\int{\frac{1}{x(logx)^m}}\,dx$ , x >0
let logx = t
differentiate both sides,
1/x dx = dt , put it above Integration.

so, $\int{\frac{1}{x(logx)^m}}\,dx=\int{\frac{dt}{t^m}}$
=$t^{-m}.dt$
= $\frac{t^{1-m}}{1-m}+C$
now, put t = logx
so, I = $\bf{\frac{(logx)^{1-m}}{1-m}}$