Results 1 to 5 of 5

Math Help - integate

  1. #1
    DBA
    DBA is offline
    Member
    Joined
    May 2009
    Posts
    129

    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}

    But the answer should be
    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Miss's Avatar
    Joined
    Mar 2010
    From
    Planet earth.
    Posts
    165
    Quote Originally Posted by DBA View Post
    \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}

    But the answer should be
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    DBA
    DBA is offline
    Member
    Joined
    May 2009
    Posts
    129
    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member Miss's Avatar
    Joined
    Mar 2010
    From
    Planet earth.
    Posts
    165
    Quote Originally Posted by DBA View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    DBA
    DBA is offline
    Member
    Joined
    May 2009
    Posts
    129
    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)
    <br />
\int \frac{1}{1-u^2} dx=\int [ \frac{1/2}{1-u}+ \frac{1/2}{1+u}]dx<br />

    Then I went on as I described in the describtion of the problem.
    Last edited by DBA; March 3rd 2010 at 10:40 AM. Reason: different notation
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Integate with arcsin x^1/2
    Posted in the Calculus Forum
    Replies: 7
    Last Post: August 24th 2010, 04:06 AM

Search Tags


/mathhelpforum @mathhelpforum