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How to integrate 1/x^2 (1 - x^2)

  1. Jan 4, 2006 #1
    May i know how to integ 1/x^2 (1-x^2) ??
    i try to form Ax/x^2 + Bx+C/(1-x^2) but can't get the answer.....


    the answer is-1/x + 1/2 ln (1+x) - 1/2 ln (1-x).pls help........
     
  2. jcsd
  3. Jan 4, 2006 #2
    use latex already!
    anyway, the form should be [tex]\frac{Ax+B}{x^2}+\frac{Cx+D}{1-x^2}[/tex]
    its alway a polynom with one order lower...
     
    Last edited: Jan 4, 2006
  4. Jan 4, 2006 #3

    VietDao29

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    Homework Helper

    In fact, you should notice that 1 - x2 = (1 - x) (1 + x).
    So partial fraction it gives:
    [tex]\frac{1}{x ^ 2 (1 - x ^ 2)} = \frac{A}{x} + \frac{B}{x ^ 2} + \frac{C}{1 - x} + \frac{D}{1 + x}[/tex].
    There are several ways to do this:
    ------------------
    The most common way is to multiply both sides by x2(1 - x2), you'll have:
    [tex]1 = Ax(1 - x ^ 2) + B(1 - x ^ 2) + Cx ^ 2 (1 + x) + Dx ^ 2 (1 - x)[/tex].
    Now plug some value of x in the RHS of the equation and solvwe for x. You'll need to plug in at least 4 values of x, since there are 4 unknowns (A, B, C, and D). You can choose the value of x such that the RHS become as simple as possible. For example, you can choose x = 0 (A, C, and D will disappear in the RHS), x = 1 (A, B, D will disappear), x = -1 (A, B, C will disappear), and choose a random x (since you'll need 4 values of x) (e.g: x = 2).
    ------------------
    The second way is to notice that the LHS of this equation:
    [tex]\frac{1}{x ^ 2 (1 - x ^ 2)} = \frac{A}{x} + \frac{B}{x ^ 2} + \frac{C}{1 - x} + \frac{D}{1 + x}[/tex]
    is an even function, so the RHS must also be an even function.
    That means:
    [tex]\frac{1}{x ^ 2 (1 - x ^ 2)} = \frac{A}{x} + \frac{B}{x ^ 2} + \frac{C}{1 - x} + \frac{D}{1 + x} = - \frac{A}{x} + \frac{B}{x ^ 2} + \frac{C}{1 + x} + \frac{D}{1 - x}[/tex].
    That means A = 0, C = D, so you are left with 2 unknowns:
    [tex]\frac{1}{x ^ 2 (1 - x ^ 2)} = \frac{B}{x ^ 2} + \frac{C}{1 - x} + \frac{C}{1 + x}[/tex].Plug in 2 values of x, and solve for B, and C.
    ------------------
    The third way (even faster) is to notice the fact that:
    [tex]\frac{1}{x ^ 2} + \frac{1}{1 - x ^ 2} = \frac{1 - x ^ 2 + x ^ 2}{x ^ 2 (1 - x ^ 2)} = \frac{1}{x ^ 2 (1 - x ^ 2)}[/tex].
    So:
    [tex]\frac{1}{x ^ 2 (1 - x ^ 2)} = \frac{1}{x ^ 2} + \frac{1}{1 - x ^ 2}[/tex].
    Just do the same for [tex]\frac{1}{(1 - x ^ 2)} = \frac{1}{(1 - x) (1 + x)}[/tex]. Can you go from here?
     
    Last edited: Jan 4, 2006
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