Answer
diverges
Work Step by Step
Given $$\int_{1}^{\infty} \frac{d x}{x^{1 / 3}+x^{2 / 3}}$$
Since for $x\geq1$
\begin{align*}
x^{1 / 3}+x^{2 / 3}&=x^{1 / 3}\left(1+x^{1 / 3}\right) \\
&\leq x^{1 / 3}\left(x^{1 / 3}+x^{1 / 3}\right)\\
&=x^{1 / 3}\left(2 x^{1 / 3}\right)\\
&=2 x^{2 / 3}\\
\frac{1}{x^{1 / 3}+x^{2 / 3}}& \geq \frac{1}{2 x^{1 / 3}}
\end{align*}
and $\int_{1}^{\infty} \frac{d x}{x^{2 / 3}} $diverges, so by the comparison test, $\int_{1}^{\infty} \frac{d x}{x^{1 / 3}+x^{2 / 3}}$ $ also diverges.
