# Math Help - integate

1. ## integate

$\int \frac{x}{x^{1/3}-x}dx$
I set u=x^{1/3} and 3u^{2} du = dx
u^3 = x

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

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

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

I make a devision and get

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

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

I do integration by parts and get
$1-u^2 = \frac{1/2}{1-u}+ \frac{1/2}{1+u}$

and the integral then

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

and my solution

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

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

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

I am cannot figure out where I go wrong.
Thanks for any help.

2. Originally Posted by DBA
$\int \frac{x}{x^{1/3}-x}dx$
I set u=x^{1/3} and 3u^{2} du = dx
u^3 = x

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

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

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

I make a devision and get

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

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

I do integration by parts and get
$1-u^2 = \frac{1/2}{1-u}+ \frac{1/2}{1+u}$

and the integral then

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

and my solution

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

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

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

I am cannot figure out where I go wrong.
Thanks for any help.
From the second line, you have $dx$ in each integral, but they were in terms of u.
I think its just a typo.
Sorry, but I did not check the long division step.
but for the integral $\int \frac{du}{1-u^2}$
You can:
1- Rewrite $1-u^2=(1-u)(1+u)$ and use partial fraction method.
2- Use a "hyperbolic" substitution.
3- evaluate it as an infinite series, since $\frac{1}{1-u^2}=\sum_{n=0}^{\infty} u^{2n}$ where $|u|<1$.

Also for the original integral itself, you can do it as an infinite series.
its easy to prove that $\frac{x}{x^{\frac{1}{3}}-x}=-\sum_{n=0}^{\infty} x^{\frac{-2n}{3}}$ where $|x|<1$.

3. Thanks for answering, but we did not have series yet. Thus, I do not understand what you did.
Is it possible to write it different without using series?
Thanks

4. Originally Posted by DBA
Thanks for answering, but we did not have series yet. Thus, I do not understand what you did.
Is it possible to write it different without using series?
Thanks
Ohhh. Sorry for that.
I told you, you can rewrite $1-u^2$ as $(1-u)(1+u)$ and use the partial fraction method.
Do you know this method?

5. Yes, I did that. I just used the wrong term. I called it "integration by parts ", sorry about that. So I did that and the solution of that wasI wrote it different, I guess the way I wrote it wasn't good)
$
\int \frac{1}{1-u^2} dx=\int [ \frac{1/2}{1-u}+ \frac{1/2}{1+u}]dx
$

Then I went on as I described in the describtion of the problem.