Given $p \neq 1$, then $\int \frac{\mathrm{d} x}{x\left(\log _{e} x\right)^{p}}$ is equal to :

$\frac{\left(\log _{e} x\right)^{1-p}}{1-p}$

p ≠ 1 , then  $I = \int \frac{dx}{x(log ~ x)^p}$

let y = log x

dy = $\frac{1}{x}$ dx

$\Rightarrow I =\int \frac{1}{y^p} d y$

$=\frac{y^{-p+1}}{-p+1} ≡ \frac{y^{1-p}}{1-p}$

$\Rightarrow I=\frac{(\log x)^{1-p}}{1-p}$