Use the Comparison Test

For any $p > 1$ choose $a$ such that $0 < a < p-1$.

We have

$$\frac{\ln x}{x^p} = \frac{\ln x^a}{ax^p} < \frac{x^a}{ax^p} = \frac{1}{ax^{p-a}}$$

Since $p-a > 1$ we have convergence for any $p > 1$.

The integral diverges if $p \leqslant 1$ by an easy comparison: $\ln x / x^p > 1/x^p$