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Math Help - I need help with partial fractions

  1. #1
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    Sep 2009
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    I need help with partial fractions

    I need to compute this integral:
    int ((1-x^2)/(x^3+x)) dx
    I think I should use the partial fraction method to simplify the fraction
    so
    (1-x^2)/(x^3+x)= A/x + B/ (1+x^2)
    Therefore
    A(1+x^2)+B(x)=1-x^2
    putting it in a polynomial form:
    x^2 (A)+x (B) + 1 (A)= 1- x^2
    and by equating the coefficients,
    A = -1
    B = 0
    However, A doesn't satisfy the constant on the LHS (+1)

    So what is the thing that I did wrong??

    Thanks in advance
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by farisallil View Post
    I need to compute this integral:
    int ((1-x^2)/(x^3+x)) dx
    I think I should use the partial fraction method to simplify the fraction
    so
    (1-x^2)/(x^3+x)= A/x + B/ (1+x^2)
    Therefore
    A(1+x^2)+B(x)=1-x^2
    putting it in a polynomial form:
    x^2 (A)+x (B) + 1 (A)= 1- x^2
    and by equating the coefficients,
    A = -1
    B = 0
    However, A doesn't satisfy the constant on the LHS (+1)

    So what is the thing that I did wrong??

    Thanks in advance

    ok

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

    if you have a denominator with a second degree the nominator of it should be a first degree not constant so

    \frac{1-x^2}{x^3+x} = \frac{A}{x} + \frac{Bx+C}{x^2+1}

    to find A,B and C

    1-x^2 = A(x^2+1) + Bx^2 + Cx

    A+B = -1
    C =0
    A=1 find B from the last one

    so you have A=1 , B=-2 ,C=0

    \frac{1-x^2 }{x^3 + x} = \frac{1}{x} + \frac{-2x}{x^2+1}

    to make sure my solution is correct

    \frac{1}{x} + \frac{-2x}{x^2+1} = \frac{x^2+1}{x^3+x} + \frac{-2x^2}{x^3+x } =\frac{1-x^2 }{x^3+x}
    and this is the left side
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  3. #3
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    ooooooo
    I see

    Thank you very much Amer, I really appreciate your help
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