Evaluating $\int_0^{2} \frac{dx}{\sqrt[3]{2x^2-x^3}}$

Question

How to calculate this integral? $$\int_0^{2} \frac{dx}{\sqrt[3]{2x^2-x^3}}$$

I suppose that it should be parted like this: $$\int_0^{1} \frac{dx}{\sqrt[3]{2x^2-x^3}} + \int_1^{2} \frac{dx}{\sqrt[3]{2x^2-x^3}}$$ but I have no idea how to calculate these two.

Thank you for help in advance.

Answer 1

$\sqrt[3]{2x^2-x^3} = x\sqrt[3]{\dfrac{2}{x}-1} \to u = \sqrt[3]{\dfrac{2}{x} - 1} \to u^3 = \dfrac{2}{x} - 1 \to x = \dfrac{2}{u^3+1}$. Can you take it from here?