I'm new with this topic and I try to solve this integral $$\int \limits^{-1}_{-\infty} \frac{x^2dx}{(4+x^3)^3}$$ There is a break point in $\displaystyle{x= -\sqrt[3]{4}\thinspace}$
I think, that I need to divide line segment into two like this: $$\int \limits^{{-\sqrt[3]{4}\thinspace}}_{-\infty} \frac{x^2dx}{(4+x^3)^3} + \int \limits^{-1}_{{-\sqrt[3]{4}\thinspace}} \frac{x^2dx}{(4+x^3)^3}$$ But here I face double limits in the first integral. How should I deal with this?
As $x\to -\infty$, $\frac{x^2}{(4+x^3)^3}\sim \frac{1}{x^4}$ and so there is no issue at the lower limit.
Next, note that
$$x^3+4=(x+4^{1/3})(x^2-4^{1/3}x+4^{2/3})$$
Hence, as $x\to -4^{1/3}$, $\frac{x^2}{(4+x^3)^3}\sim \frac{4^{2/3}}{432(x+4^{1/3})^3}$ and the integral diverges.