Answer
Divergent
Work Step by Step
Let \[I=\int_{0}^{\infty}\frac{x^2}{\sqrt{1+x^3}}dx\;\;\;\ldots(1)\]
\[I=\lim_{t\rightarrow \infty}\int_{0}^{t}\frac{x^2}{\sqrt{1+x^3}}dx\;\;\;\ldots (2)\]
\[I=\lim_{t\rightarrow \infty}\frac{1}{3}\int_{0}^{t}\frac{3x^2}{\sqrt{1+x^3}}dx\;\;\;\ldots (3)\]
Let\[I_1=\int\frac{3x^2}{\sqrt{1+x^3}}dx\]
Substitute $r=1+x^3$ _____(4)
$\;\;\;\;\;\;\;\;\;\Rightarrow dr=3x^2dx$
\[I_1=\int\frac{dr}{\sqrt{r}}=2\sqrt{r}\]
Using (4)
\[I_1=2\sqrt{1+x^3}\;\;\;\ldots (5)\]
Using (5) in (3)
\[I=\lim_{t\rightarrow \infty}\frac{1}{3}\left[2\sqrt{1+x^3}\right]_{0}^{t}\]
\[I=\lim_{t\rightarrow \infty}\left[\frac{2}{3}\sqrt{1+t^3}-\frac{2}{3}\right]\]
Does not exist
Since limit on R.H.S. of (2) does not exist so $I$ is divergent.
