Convergence of the integral $\int_0^\infty \frac{\ln^3(1+x^{1/4})}{x^{1/5}+x^2}\,\arctan(x)\;dx$ How to prove that this integral is convergent?
$$\int_0^\infty \frac{\ln^3(1+x^{1/4})}{x^{1/5}+x^2}\,\arctan(x)\;dx$$
I have a little experience with this kind of problem, I know we should solve the problem for the integral from $0$ to $1$ and from $1$ to $\infty$, but I couldn't find good examples so I'm not sure how to proceed.
 A: What follows is a rather rough outline.
Split the integral up: $$\int_0^\infty \frac{\ln^3(1+x^{1/4})}{x^{1/5}+x^2}\,\arctan(x)\;dx=\int_0^1\frac{\ln^3(1+x^{1/4})}{x^{1/5}+x^2}\,\arctan(x)\;dx+\int_1^\infty \frac{\ln^3(1+x^{1/4})}{x^{1/5}+x^2}\,\arctan(x)dx$$
Use L'Hôpital's rule to show that: $$\lim_{x \to 0^+}\frac{\ln^3(1+x^{1/4})}{x^{1/5}+x^2}\,\arctan(x)=0$$
Hence the left integrand can be defined to equal $0$ at $x=0$, hence the integrand is continuous and therefore has a finite integral in the closed interval $[0,1]$.
The second integral is a bit trickier. The term $\arctan(x)$ is irrelevant because it can be bounded by $\pi /2$, so I'll ignore it. 
Recall that the natural logarithm has  smaller order of magnitude than $f(x)=x^{1/3}$, in particular there exists $M>1$ such that:
$$t ≥ M \implies \ln(t)<t^{1/3}$$
Note that $x>M^4$ implies $1+x^{1/4}>M$, which gives $\ln^3(1+x^{1/4})<1+x^{1/4}$, hence:
$$
\begin{align}
&\int_1^\infty \frac{\ln^3(1+x^{1/4})}{x^{1/5}+x^2}dx \\
&\leq
\int_1^{M^4}\frac{\ln^3(1+x^{1/4})}{x^{1/5}+x^2}dx+\int_{M^4}^\infty\frac{\ln^3(1+x^{1/4})}{x^{1/5}+x^2}dx \\ &\leq
\int_1^{M^4}\frac{\ln^3(1+x^{1/4})}{x^{1/5}+x^2}dx+\int_{M^4}^\infty\frac{1+x^{1/4}}{x^{1/5}+x^2}dx
\end{align}
$$
The first integral is clearly finite. To see that the second one is also finite, note that, splitting the numerator: $$\int_{M^4}^\infty\frac{1+x^{1/4}}{x^{1/5}+x^2}dx≤\int_{M^4}^\infty\frac{1}{x^2}dx+\int_{M^4}^\infty\frac{x^{1/4}}{x^2}dx≤\int_{M^4}^\infty\frac{1}{x^2}dx+\int_{M^4}^\infty\frac{1}{x^{7/4}}dx$$
Both of these integrals converge.
