Integrate the function $\frac{1}{x(\log x)^{m}}$,where $x > 0$ and $m \neq 1$.

(N/A) Let $\log x = t$.
Then,differentiating both sides with respect to $x$,we get $\frac{1}{x} dx = dt$.
Substituting these into the integral,we have:
$\int \frac{1}{x(\log x)^{m}} dx = \int \frac{1}{t^{m}} dt$.
This can be written as $\int t^{-m} dt$.
Using the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$),we get:
$\frac{t^{-m+1}}{-m+1} + C = \frac{t^{1-m}}{1-m} + C$.
Substituting back $t = \log x$,the final result is:
$\frac{(\log x)^{1-m}}{1-m} + C$,where $C$ is an arbitrary constant.