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Math Help - Integrate

  1. #1
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    Integrate

    what is the easiest way to integrate this

    \frac{-14x+18x^3}{(3x^2-1)(1-x^2)}

    thanks
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by charikaar View Post
    what is the easiest way to integrate this
    By partial fraction decompostion.
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  3. #3
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    is this correct

     \frac{-14x+18x^3}{ (\sqrt{3}x - 1)(\sqrt{3}x+1)(1+x)(1-x)} = \frac{2\sqrt3}{\sqrt{3}x - 1}+\frac{-2\sqrt3}{\sqrt{3}x + 1} + \frac{1}{1-x} + \frac{-1}{1+x}
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by charikaar View Post
    what is the easiest way to integrate this

    \frac{-14x+18x^3}{(3x^2-1)(1-x^2)}

    thanks
    I do not know if my way is the easiest but here it is


    \frac{18x^3 -14x }{(3x^2-1)(1-x^2)}

    \left(\frac{18x^3-14x}{(3x^2-1)}\right)\left(\frac{1}{1-x^2}\right)

    \left(6x - \frac{8x}{(3x^2-1)} \right) \left(\frac{1}{1-x^2} \right)

    \int \frac{6x}{1-x^2 } - \frac{8x}{(3x^2-1)(1-x^2)} \cdot dx

    sub x^2 = u \Rightarrow 2x \cdot dx = du \Rightarrow dx = \frac{du}{2x}

     \int \frac{3}{1-u} - \frac{4}{(3u-1)(1-u)} \cdot du

    second one you can integrate it by partial fraction first one is easy
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by charikaar View Post
    is this correct

     \frac{-14x+18x^3}{ (\sqrt{3}x - 1)(\sqrt{3}x+1)(1+x)(1-x)} = \frac{2\sqrt3}{\sqrt{3}x - 1}+\frac{-2\sqrt3}{\sqrt{3}x + 1} + \frac{1}{1-x} + \frac{-1}{1+x}
    I tried to check it for you, but I think that there is an easier way.

    Note that

    \frac{18x^3-14x}{(3x^2-1)(x-1)(x+1)}=\frac{Ax+B}{3x^2-1}+\frac{C}{x-1}+\frac{D}{x+1}

    18x^3-14x=(Ax+B)(x^2-1)+C(3x^2-1)(x+1)+D(3x^2-1)(x-1)

    x=1: 4=4C\Rightarrow{C}=1

    x=-1: -4=-4D\Rightarrow{D}=1

    Now With the values of C and D known, we put them in and then

    x=0: 0=-B-1+1\Rightarrow{B}=0

    And now use any arbitrary value of x to solve for A.

    x=2\Rightarrow{A}=12 s.t.

    \frac{18x^3-14x}{(3x^2-1)(x-1)(x+1)}=\frac{12x}{3x^2-1}+\frac{1}{x-1}+\frac{1}{x+1}
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