Convergence of $\int_0^{+\infty}\frac{\ln(x^{2\over3}+2)|\ln^{\alpha-{2\over3}}(x+x^2)|}{(x^{1\over3}+1)^2x^{\alpha}}$,depending on $\alpha$

Question

The question asks me for what values of $\alpha \in\mathbb{R}$ this integral converges.Firstly,i split the domain: $$\int_0^{+\infty}\frac{\ln(x^{2\over3}+2)|\ln^{\alpha-{2\over3}}(x+x^2)|}{(x^{1\over3}+1)^2x^{\alpha}}=\int_0^2\frac{\ln(x^{2\over3}+2)|\ln^{\alpha-{2\over3}}(x+x^2)|}{(x^{1\over3}+1)^2x^{\alpha}}+\int_2^{+\infty}\frac{\ln(x^{2\over3}+2)|\ln^{\alpha-{2\over3}}(x+x^2)|}{(x^{1\over3}+1)^2x^{\alpha}}$$ Then integral is well defined for $x=2$ and I have to consider behavior of my function when $x\to0^+$ and $x\to+\infty$.For $x\to+\infty$ :$$\int_2^{+\infty}\frac{\ln(x^{2\over3}+2)|\ln^{\alpha-{2\over3}}(x+x^2)|}{(x^{1\over3}+1)^2x^{\alpha}}\sim \int_2^{+\infty}\frac{\ln(x^{2\over3})|\ln^{\alpha-{2\over3}}(x^2)|}{x^{\alpha+{2\over3}}}$$.And we know it's convergent when $\alpha+{2\over3}\gt1\implies\alpha\gt{1\over3}$(Because $\int_2^{+\infty}\frac{1}{x^\beta\ln^\gamma}$ is convergent $\forall\beta\gt1$,irrespective of $\gamma$)
As for $x\to0^+$: $$\int_0^2\frac{\ln(x^{2\over3}+2)|\ln^{\alpha-{2\over3}}(x+x^2)|}{(x^{1\over3}+1)^2x^{\alpha}}\sim\int_0^2\frac{\ln(2)|\ln^{\alpha-{2\over3}}(x)|}{x^{\alpha}}$$For $\alpha={2\over3}$ we have $\int_0^2\frac{\ln(2)}{x^{2\over3}}$ And this is convergent because $\int_0^{b\in\mathbb{R}}\frac{C}{x^\epsilon}$ is convergent $\forall\epsilon\lt1$ and constant $C$.But i can't proceed and determine for what $\alpha$ exactly this integral is convergent.