# integate

• Mar 3rd 2010, 09:35 AM
DBA
integate
$\displaystyle \int \frac{x}{x^{1/3}-x}dx$
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}

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

I am cannot figure out where I go wrong.
Thanks for any help.
• Mar 3rd 2010, 09:50 AM
Miss
Quote:

Originally Posted by DBA
$\displaystyle \int \frac{x}{x^{1/3}-x}dx$
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}

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

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

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

Also for the original integral itself, you can do it as an infinite series.
its easy to prove that $\displaystyle \frac{x}{x^{\frac{1}{3}}-x}=-\sum_{n=0}^{\infty} x^{\frac{-2n}{3}}$ where $\displaystyle |x|<1$.
• Mar 3rd 2010, 09:59 AM
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
• Mar 3rd 2010, 10:05 AM
Miss
Quote:

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. :o
I told you, you can rewrite $\displaystyle 1-u^2$ as $\displaystyle (1-u)(1+u)$ and use the partial fraction method.
Do you know this method?
• Mar 3rd 2010, 10:28 AM
DBA
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 was:(I wrote it different, I guess the way I wrote it wasn't good)
$\displaystyle \int \frac{1}{1-u^2} dx=\int [ \frac{1/2}{1-u}+ \frac{1/2}{1+u}]dx$

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