Complicated Improper Integral convergence/divergence Does the following integral converge for  $x < 0$
$$\int _x^0\:\cfrac{\ln^2 \ (|t |)}{ \sqrt[3]{t} }dt$$
I tried splitting it into two separate integrals under the assumption that both are convergent:
$$\int _x^{-1}\:\cfrac{\ln^2 \ (|t |)}{ \sqrt[3]{t} }dt + \int _{-1}^{0}\:\cfrac{\ln^2 \ (|t |)}{ \sqrt[3]{t} }dt$$
By comparison test:
$$\bigg| \int _x^{-1}\:\cfrac{\ln^2 \ (|t |)}{ \sqrt[3]{t} }dt \bigg| \leq \int _x^{-1}\:{\ln^2 \ (|t|)} dt $$
this part is convergent.
The problem is ,  I cannot find a way to show it for the other part:
$$\bigg| \int _{-1}^{0}\:\cfrac{ln^2 \ (|t |)}{ \sqrt[3]{t} }dt \bigg| \leq \int _{-1}^{0}\textit{something} $$
Is this even the right approach?
 A: The function $t\mapsto \frac{\ln^2\left(\lvert t\rvert\right)}{t^{1/3}}$ is continuous over $(-\infty, 0)$ so the only possible issue with your integral is at 0. If we show that, for example
$$\int _{-1}^0\:\cfrac{\ln^2 \ (|t |)}{ t^{1/3} }dt$$
converges then your problem is solved. To make things more practical, let's go back to positive numbers. Let $\varepsilon\in(0,1)$ and write the change of variable $u=-t$
$$\int _{-1}^{-\varepsilon}\cfrac{\ln^2 \ (|t |)}{ t^{1/3} }dt = \int _{-1}^{-\varepsilon}\cfrac{\ln^2 \ (-t)}{ t^{1/3} }dt = -\int_{\varepsilon}^1\frac{\ln^2u}{u^{1/3}}du$$
Letting $\varepsilon\rightarrow 0$, your problem is thus equivalent to showing the convergence of
$$\int_{0}^1\frac{\ln^2u}{u^{1/3}}du$$
Write one more change of variable $v=\ln(u)$
$$\int_{\varepsilon}^1\frac{\ln^2u}{u^{1/3}}du = \int_{\ln \varepsilon}^0e^{\frac{2v}{3}}v^2dv$$
Letting $\varepsilon\rightarrow 0$ again, we find
$$\int_{0}^1\frac{\ln^2u}{u^{1/3}}du = \int_{-\infty}^0e^{\frac{2v}{3}}v^2dv=\frac{27}{4}$$
This method also allows you to compute your integral modulo some work on the integration boundaries.
A: Assume $t>0$ and let $t=u^3$ to make
$$\int _x^0\:\cfrac{\ln^2 \ (|t |)}{ \sqrt[3]{t} }dt=27\int_{\sqrt[3]x}^0 u \log^2(u)\,du=-\frac{3x^{2/3}}{4} \, (2 (\log (x)-3) \log (x)+9)$$
Assume $t<0$ and let $t=-u^3$
