Results 1 to 2 of 2

Thread: Indefinite integral

  1. #1
    Ant
    Ant is offline
    Member
    Joined
    Apr 2008
    Posts
    145
    Thanks
    4

    Indefinite integral

    Hi,

    I'm trying to integrate:

    $\displaystyle 1/(ax - x^3)$

    I've tried using partial fractions; and I've got (1/a) lnx and a^(-3/2) tanh(x/a).

    Thanks for any help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    substitute $\displaystyle x=\dfrac 1t$ and the integral becomes $\displaystyle \displaystyle-\int{\frac{t}{a{{t}^{2}}-1}\,dt}=-\frac{1}{2a}\ln \left| 1-a{{t}^{2}} \right|+k,$ so back substitute and get $\displaystyle -\dfrac{1}{2a}\ln \left| \dfrac{{{x}^{2}}-a}{{{x}^{2}}} \right|+k.$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Indefinite integral
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Jul 28th 2010, 09:30 AM
  2. indefinite integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Mar 1st 2010, 10:16 AM
  3. Indefinite Integral
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Feb 22nd 2010, 06:44 PM
  4. Indefinite Integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Nov 19th 2007, 10:56 AM
  5. indefinite integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Nov 9th 2007, 02:23 PM

Search Tags


/mathhelpforum @mathhelpforum