Answer
The integral $\int_{1}^{\infty} \dfrac{dx}{\sqrt {x^{1/3}+x^3}}$ converges.
Work Step by Step
We are given the function
$f(x)=\int_{1}^{\infty} \dfrac{dx}{\sqrt {x^{1/3}+x^3}}$
Since, $\sqrt {x^{1/3}+x^3} \geq \sqrt {x^3}$
$\dfrac{dx}{\sqrt {x^{1/3}+x^3}} \leq \dfrac{1}{x^{3/2}} $
But the integral $\int_{1}^{\infty} \dfrac{dx}{x^{3/2}} $ shows a p-type integral with $\dfrac{3}{2} \gt 1$. Thus, the integral $\int_{1}^{\infty} \dfrac{dx}{x^{3/2}}$ converges.
Therefore, by the comparison test, the integral $\int_{1}^{\infty} \dfrac{dx}{\sqrt {x^{1/3}+x^3}}$ converges as well.
