Without using beta function, for what $p$ does $\int_0^\infty \frac{\log(x)}{1+x^p} dx$ converge? Question:

For what $p$ does $\int_0^\infty \frac{\log(x)}{1+x^p} dx$ converge?

Is there a simpler solution to this integral than approaching it via the beta function (which I have not yet learned but have seen it solved this way)? Specifically, can someone put forward a solution that looks at the integrand at its "bad" areas where it might fail to converge, thus causing the entire integral to fail to converge?
First, I notice that if $p>0$ then $x^p \approx 0$ near $x=0$. Or if $p<0$, then $x^p \to \infty$ around $x=0$.
 A: First consider the integral over $[0,1]$.
For all $p \in \mathbb{R}$,
$$-\frac{\log(x)}{1 + x^p}\leqslant -\log(x),$$
and 
$$\int_0^1 -\log(x) \,dx=\lim_{\epsilon \to 0}\int_\epsilon^1 -\log(x) \,dx = \lim_{\epsilon \to 0}[x-x \log(x)]|_\epsilon^1 \\= \lim_{\epsilon \to 0}[1 - \epsilon+ \epsilon\log(\epsilon)]= 1.$$
Therefore, the integral over $[0,1]$converges for all $p \in \mathbb{R}$ by the comparison test .
The integral over $[1,\infty)$ converges for $p > 2$ by the comparison test since
$$\frac{\log(x)}{1 + x^p}\leqslant x^{-(p-1)},$$
and $\int_1^{\infty}x^{-(p-1)}\,dx$ converges for $p > 2$.
Also,
$$\lim_{x \to \infty}\frac{\log(x)}{(1+x^p)x^{-p}}= \lim_{x \to \infty}\frac{\log(x)}{1+x^{-p}}= \infty.$$
By the limit comparison test, the integral over $[1,\infty)$ diverges for $p \leqslant 1$ since $\int_1^{\infty}x^{-p}\,dx$ diverges for $p \leqslant 1.$
For $1 < p \leq 2$, use
$$\frac{\log(x)}{2x^p} \leqslant \frac{\log(x)}{1+x^p} \leqslant \frac{\log(x)}{x^p},$$
and compare with
$$\int_1^{\infty} \frac{\log(x)}{x^p} \, dx = \int_0^{\infty} xe^{(1-p)x} \, dx$$
which converges for $p > 1$.
