# A difficult integral?

1. Nov 24, 2009

### niz73

Hi all,

2
∫ [x (8-x3)^1/3 ] dx
0

Last edited: Nov 24, 2009
2. Nov 24, 2009

### Pengwuino

What have you attempted for the problem so far?

3. Nov 24, 2009

### niz73

I have substituted u^3 = 8 - x^3 then 3 u^2 du = - 3 x^2 dx
But now the problem has only x so in order to substitute for dx I have to divide and multiply by x and that means I can not eliminate cube root.