Substitute \(u=\ln x\), so that \(\mathrm du=\dfrac{\mathrm dx}x\). The integral is then equivalent to
\(\displaystyle\int\frac{\mathrm dx}{x\ln^px}=\int\frac{\mathrm du}{u^p}=\begin{cases}\dfrac{u^{p+1}}{p+1}+C&\text{for }p
eq1\\\\\ln|u|+C&\text{for }p=1\end{cases}\)
Then transforming back to \(x\) gives
\(\displaystyle\int\frac{\mathrm dx}{x\ln^px}=\begin{cases}\dfrac{\ln^{p+1}x}{p+1}+C&\text{for }p
eq1\\\\\ln|\ln x|+C&\text{for }p=1\end{cases}\)
