Find if the integral $ \int_{0}^{\infty}\frac{\tan^{-1}(x)}{(2x+1)(x^4-6x^3+9x^2)^{1/3}}\,dx$ converges without solving it I'm trying to solve this integral, but without any success.
Is there a way to tell if the integral is convergent or divergent without actually solving it? Am I missing something?
$$\int_{0}^{\infty} \dfrac{\arctan x}{(2x+1)(x^4-6x^3+9x^2)^{1/3}}\,dx$$

We want to show $$\int_{0}^{\infty} \dfrac{\arctan x}{(2x+1)(x^4-6x^3+9x^2)^{1/3}}\,dx$$ converges. We'll consider an initial and final segment separately and bound them both.
Let's look at the factors. For large $x$ we have
$$\arctan(x) \le \pi/2 \qquad \dfrac{1}{(2x+1) } \simeq \frac{1}{2x} \qquad \dfrac{1}{ (x^4-6x^3+9x^2)^{1/3}} \simeq \frac{1}{x^{4/3}}$$
. . . since in the last factor the $x^4$ dominates.
Make the above formal and use it to compare some final segment of the integral to
$$\int_{x}^{\infty} \dfrac{\pi/2}{(2x )x^{4/3}}\,dx = \frac{\pi}{2}\int_{x}^{\infty} \dfrac{dx}{ x^{7/3}} $$
For small $x$ we have
$$\arctan(x) \simeq  x -\frac{x^3}{3}+\frac{x^5}{5} - \ldots \qquad \dfrac{1}{(2x+1) } \simeq \frac{1}{2} \qquad \dfrac{1}{ (x^4-6x^3+9x^2)^{1/3}} \simeq \dfrac{1}{ ( 9x^2)^{1/3}}$$
. . .  since in the last factor the $x^2$ term dominates.
Use the above similar to before to bound some integral over an initial segment.
Now consider the integral near $3$ where it blows up:

This time write $ x^4-6x^3+9x^2= x^2 (x-3)^2 $ and so
$$ \dfrac{1}{ (x^4-6x^3+9x^2)^{1/3}}  = \frac{1}{x^{2/3} (x-3)^{2/3}}$$
Since everything else is bounded you only need to show the integral of $\frac{1}{ (x-3)^{2/3}}$ around $3$ converges.
We conclude the integrals at $0,3$ and $\infty$ all converge. Since the function is continuous everywhere else it is bounded over the remaining domain hence has finite integral.
