The title pretty much sums up the problem. I need to examine for which values of $p\in\mathbb{R}$ the following improper integral converges: $$I := \int_1^\infty\frac{1}{\left(x^{p^3} + x^{4p}\right)\ln^2(1+\sqrt[4]{x})}dx$$
What I initially thought about was the Abel-Dirichlet test, which only gives us a range of $p$ for which the integral converges, but I'm not sure how to find out whether that are all values and I even suspect that they're not. So, if we define $$f(x) := \frac{1}{\ln^2(1+x^{\frac{1}{4}})}$$ then we have that $f(x) > 0$ and is monotonically decreasing and $\displaystyle\lim_{x\to\infty}f(x) = 0$, so if we make $\displaystyle\int_1^{\infty}\frac{1}{x^{p^3}+x^{4p}}dx$ convergent, then by the Abel-Dirichlet test $I$ will also be convergent. And this happens when $max\{p^3, 4p\} > 1$, i.e. when $p > \frac{1}{4}$.
But as said, this is an artificial constraint in order to employ the test I could think of. I also figured I could somehow bound $I$ between some integrals that converge for the same values of $p$, but I could only find the upper bound in the following way by examining boundedness of $f(x)$:
Since we mentioned it was monotonically decreasing and tends to $0^+$, from some point on it will be strictly less than 1. This point is namely $a = (e-1)^4 > 1$, so we can split $I$ as sum of two integrals (one from 1 to a and another from a to $\infty$). Since the first one is a definite integral, it is a number and doesn't change convergence, so $$I \sim \int_a^\infty\frac{1}{\left(x^{p^3} + x^{4p}\right)\ln^2(1+\sqrt[4]{x})}dx < \int_a^\infty\frac{1}{x^{p^3} + x^{4p}}dx$$ and we obtain the same results as with the test before, namely that when $p > \frac{1}{4}$ the upper bound integral converges and so does $I$. But what could be a smart lower bound? Is that even the approriate approach? Any ideas are greatly welcomed! Many thanks in advance!
