Substitute [tex]u=\ln x[/tex], so that [tex]\mathrm du=\dfrac{\mathrm dx}x[/tex]. The integral is then equivalent to
[tex]\displaystyle\int\frac{\mathrm dx}{x\ln^px}=\int\frac{\mathrm du}{u^p}=\begin{cases}\dfrac{u^{p+1}}{p+1}+C&\text{for }p\neq1\\\\\ln|u|+C&\text{for }p=1\end{cases}[/tex]
Then transforming back to [tex]x[/tex] gives
[tex]\displaystyle\int\frac{\mathrm dx}{x\ln^px}=\begin{cases}\dfrac{\ln^{p+1}x}{p+1}+C&\text{for }p\neq1\\\\\ln|\ln x|+C&\text{for }p=1\end{cases}[/tex]
