For what $p$ is $\frac{1}{(x(1+\ln(x)^2))^p}$ Lebesgue integrable? I'm trying to use the fact that given $f:[a,\infty)\to\mathbb{R}$ Riemann integrable for every closed interval $[c,d]\subset [a,\infty)$, then $f$ is Lebesgue integrable if, and only if, $\int_a^\infty|f(x)| \, dx$ exists.
In particular, $f(x)=\frac{1}{x(1+\ln(x)^2)}$ is $p$-Lebesgue integrable if $\int_0^\infty\frac{1}{x^p(1+\ln(x)^2)^p} \, dx<\infty $. Here using the fact that $f>0$.
But I can't solve this Riemann integral.

Using the inequality $\frac{x}{x+1} < \log (1+x) < x$ for ll $x>-1$ and $x \ne 0$. Let $x=u-1$ then the inequality becomes $\frac{u-1}{u} < \log (u) < u-1$ for all $u>0$. Thus, $\frac{(x-1)^2}{x^2}< (\ln x)^2<(x-1)^2$ so that $1+\frac{(x-1)^2}{x^2}<1+ (\ln x)^2<1+(x-1)^2$ which implies that  $\frac{1}{1+\frac{(x-1)^2}{x^2}}>\frac{1}{1+ (\ln x)^2}>\frac{1}{1+(x-1)^2}$.
Thus, $\frac{x^{2p}}{(x^2+(x-1)^2)^p}>\frac{1}{(1+ (\ln x)^2)^p}>\frac{1}{(1+(x-1)^2)^p}$.
Alos, since $x>0$ then, $$\frac{1}{x^p(1+ (\ln x)^2)^p}<\frac{x^{p}}{(x^2+(x-1)^2)^p}.$$
Using Maple 12, I tried to integrate $\int_0^{\infty}{\frac{x^{p}}{(x^2+(x-1)^2)^p}dx}$ I get that the integral converges when $p\ge2$, diverges when $p<2$.
