I'm reviewing improper integrals a bit and claim that$$\int\_0^{\infty} \frac{dx}{\sqrt[3]{x^4+x^2}}$$

converges. First off all $$\int\_0^{\infty} \frac{dx}{\sqrt[3]{x^4+x^2}}\leq\int\_0^{\infty} \frac{dx}{x^{4/3}}$$

By the integral test ; the integral on the right above converges if and only if the series $$\sum\_{k=0}^{\infty}\frac{1}{k^{4/3}}$$
converges, which it does since it is a harmonic series with $p=4/3 > 1$. The original now converges by the comparison test. Is this sound? Also, how can analyze the value of this integral?