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Math Help - Integrate this using partial fraction

  1. #1
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    Integrate this using partial fraction

    how do i integrate this

    \frac{1}{(3x^2-1)(1-x^2)}

    I have to split this into partial fraction but I don't if this is correct

    \frac{1}{(3x^2-1)(1-x^2)} =\frac{A}{(1-x^2)} + \frac{B}{3x^2-1}
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  2. #2
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    Quote Originally Posted by charikaar View Post
    how do i integrate this

    \frac{1}{(3x^2-1)(1-x^2)}

    I have to split this into partial fraction but I don't if this is correct

    \frac{1}{(3x^2-1)(1-x^2)} =\frac{A}{(1-x^2)} + \frac{B}{3x^2-1}
    No.

    Factorise the numerator into linear factors and then decompose.
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  3. #3
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    Quote Originally Posted by charikaar View Post
    how do i integrate this

    \frac{1}{(3x^2-1)(1-x^2)}

    I have to split this into partial fraction but I don't if this is correct

    \frac{1}{(3x^2-1)(1-x^2)} =\frac{A}{(1-x^2)} + \frac{B}{3x^2-1}
    Note that  1-x^2 = (1+x)(1-x) , and that, similarly,  3x^2 - 1 can be written as  (\sqrt{3}x)^2 -1 , which can be written  (\sqrt{3}x - 1)(\sqrt{3}x+1)

    Knowing this, you can write:

     \frac{1}{ (\sqrt{3}x - 1)(\sqrt{3}x+1)(1+x)(1-x)} = \frac{A}{\sqrt{3}x - 1}+\frac{B}{\sqrt{3}x + 1} + \frac{C}{1-x} + \frac{D}{1+x}

    When you perform partial fractions, functions like these are extremely easy to integrate using the following rule:

     \displaystyle \int \frac{dx}{ax+b} = \frac{1}{a} \ln(ax+b) + C
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