<s>[INST] Does this integral converge? $\int \limits^{-1}_{-\infty} \frac{x^2dx}{(4+x^3)^3}$ 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?
 [/INST] 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.
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