I have to show the following diverges: $$\int_0^1 \frac{1}{x^{1/3} -x^{4/3}} dx$$
I am meant to do this without evaluating the integral. I know that I have to split it into:
$$\int_0^{0.5} \frac{1}{x^{1/3} -x^{4/3}} dx + \int_{0.5}^1 \frac{1}{x^{1/3} -x^{4/3}} dx$$
I know that the one on the left converges and the one on the right diverges. How do I prove these two facts? I'm not sure but it may have to do with using direct comparison.
$$ \frac{1}{x^{1/3}-x^{4/3}}=\frac{1}{x^{1/3}(1 -x)}. $$ If $0\le x\le1/2$ then $$ \frac{1}{x^{1/3}(1-x)}\le\frac{2}{x^{1/3}}. $$ If $1/2\le x\le1$ then $$ \frac{1}{x^{1/3}(1-x)}\ge\frac{1}{1-x}. $$