Recall that the logarithm function satisfies the inequality
$$\log(x)\le x-1<x \tag 1$$
for $x>0$.
Then, since $\log(x^{\alpha})=\alpha \log(x)$, we have from $(1)$ for $\alpha>0$
$$\begin{align}
\alpha \log(x)&< x^{\alpha}\\\\
\log(x)&<\frac{x^{\alpha}}{\alpha} \\\\
\frac{\log(x)}{x^p}&< \frac{x^{\alpha-p}}{\alpha}
\end{align}$$
Therefore, for any $p>1$ we can take $\alpha>0$ small enough so that $p-\alpha>1$. Therefore the integral of interest converges for all $p>1$ by the comparison test.
Since $\log(x)>1$ for $x>e$, then $\frac{\log(x)}{x^p}>\frac{1}{x^p}$ for $x>e$ and the integral diverges for $p\le 1$.
We can evaluate the improper integral applying integration by parts with $u=\log(x)$ and $v=x^{1-p}/(1-p)$. Proceeding accordingly, we obtain
$$\begin{align}
\lim_{L\to \infty}\int_1^L \frac{\log(x)}{x^p}\,dx&=\lim_{L\to \infty}\left(\left. \frac{x^{1-p}\log(x)}{1-p}\right|_{1}^{L}\right)+\frac{1}{p-1}\lim_{L\to \infty }\int_1^L\frac{1}{x^p}\,dx\\\\
&=\frac{1}{(p-1)^2}
\end{align}$$