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Math Help - integration

  1. #1
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    Question integration

    i was wondering how to integrate

    3/(2x - 4x^3)

    im struggling to integrate this..are there steps to show this?

    thanks
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by dopi View Post
    i was wondering how to integrate

    3/(2x - 4x^3)

    im struggling to integrate this..are there steps to show this?

    thanks
    \frac{3}{2x-4x^3}=\frac 32 \cdot \frac{1}{x(1-2x^2)}=\frac 32 \frac{1}{x(1-\sqrt{2} x)(1+\sqrt{2} x)}

    Do you know how to use partial fraction decomposition ? Partial-Fraction Decomposition: General Techniques
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  3. #3
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    Question still cant do it..slight mistake with equation

    soory i made a slight mistake with the equation the correct one is below

    3/(2x - 4x^4)

    im still struggling with breaking it down into partial fractions
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by dopi View Post
    soory i made a slight mistake with the equation the correct one is below

    3/(2x - 4x^4)

    im still struggling with breaking it down into partial fractions
    2x-4x^4=2x(1-2x^3)=2x(1-\sqrt[3]{2}x)(1+\sqrt[3]{2}x+\sqrt[\frac{2}{3}]{x})
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  5. #5
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    \frac{1}{x\left( 1-2x^{2} \right)}=\frac{1-2x^{2}+2x^{2}}{x\left( 1-2x^{2} \right)}=\frac{1}{x}+\frac{2x}{1-2x^{2}}.

    Ahhh, well, then having \int{\frac{dx}{2x-4x^{4}}}, put x=\frac1u and you'll get an easy integral.
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