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Integrate 1/ x^2 (1-x) (1-x^2)?

  1. Jan 3, 2006 #1
    how to integ 1/ x^2 (1-x) (1-x^2)??

    the answer is
    -1/x + (1/2) ln (1+x) -(1/2) ln (1-x)
  2. jcsd
  3. Jan 3, 2006 #2
    please learn to use latex, did you mean to integrate this: [tex]\frac{1}{x^2(1-x)(1-x^2)}[/tex]?

    (click on it to see how i wrote it...)
  4. Jan 3, 2006 #3
    ya,this is what i mean
  5. Jan 3, 2006 #4
    try splitting the exprassion: [tex]\frac{1}{x^2(1-x)^2(1+x)}=\frac{1}{x^2(1-x)^2}-\frac{1}{x(1-x)^2(1+x)}[/tex]
    for example... you can keep simplifying it.
  6. Jan 3, 2006 #5
    its easy to get rid of x (put 1+x-x), the generic way to simplify is [tex]\frac{1}{(1+x)(1-x^2)}=\frac{A}{1+x}+\frac{Bx+C}{1-x^2}[/tex]
    and A B and C for them to sattisfy the equation.
  7. Jan 3, 2006 #6


    User Avatar

    Simplify the bottom to,
    1/ x^2(x-1)^2(x+1)
    1 = Ax/x^2 + Cx+D/(x-1)^2 + E/(x+1)
    1= Ax(x-1)^2(x+1) + (Cx+D)x^2(x+1) + E(x-1)^2x^2

    Collect like terms and solve for A -> E and then integrate.
  8. Jan 4, 2006 #7
    oh........okok thanx very much
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