= Solution
Interpret the cube root as the real cube root. The curved boundary is $y=-x^3$, so the bounded region is $0\le x\le3$, $-x^3\le y\le0$. In particular $y$ is negative inside it. The <double integral> is the convergent <improper integral>
$$
I=\int_0^3\frac1{x^3+1}\left(\int_{-x^3}^0y^{-1/3}\,dy\right)dx.
$$
For negative real $y$, an <antiderivative> is $(3/2)y^{2/3}=(3/2)|y|^{2/3}$. The inner <integral> is therefore $-3x^2/2$. Hence
$$
I=-\frac32\int_0^3\frac{x^2}{x^3+1}\,dx
=-\frac12\bigl[\ln(x^3+1)\bigr]_0^3,
\qquad \boxed{I=-\frac12\ln28.}
$$
The negative sign is essential. Absolute convergence follows by the same calculation with the sign removed, so the singular edge at $y=0$ causes no ambiguity.
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