$\displaystyle \int \frac{x}{x^{1/3}-x}dx$

I set u=x^{1/3} and 3u^{2} du = dx

u^3 = x

$\displaystyle \int \frac{u^3}{u-u^3} * 3u^{2}dx$

$\displaystyle \int \frac{u^3}{u-u^3}* 3u^{2}dx$

$\displaystyle 3\int \frac{u^4}{1-u^2}dx$

I make a devision and get

$\displaystyle 3\int (-u^{2}-1+ \frac{1}{1-u^2})dx$

With the last term $\displaystyle 1-u^2$ I am not sure what to do.

I do integration by parts and get

$\displaystyle 1-u^2 = \frac{1/2}{1-u}+ \frac{1/2}{1+u}$

and the integral then

$\displaystyle 3\int (-u^{2}-1+ \frac{1/2}{1-u}+ \frac{1/2}{1+u})dx$

and my solution

$\displaystyle 3[-\frac{1}{3}u^{3}-u -\frac{1}{2}ln|1-u|+\frac{1}{2}ln|1+u|)+C$

And then I would put back u=x^{1/3}

But the answer should be

$\displaystyle 3[-\frac{1}{3}u^{3}-u +\frac{1}{2}ln|-1-u|-\frac{1}{2}ln|-1+u|)+C$

I am cannot figure out where I go wrong.

Thanks for any help.