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Thread: Indefinite integral

  1. February 13th 2011, 11:52 AM #1
    Ant
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    Indefinite integral

    Hi,

    I'm trying to integrate:

    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!
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  2. February 13th 2011, 12:04 PM #2
    Krizalid
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    substitute x=\dfrac 1t and the integral becomes \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 -\dfrac{1}{2a}\ln \left| \dfrac{{{x}^{2}}-a}{{{x}^{2}}} \right|+k.
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