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How to integ 2x^2 / (1-x^2) using partial fraction?

  1. Jan 3, 2006 #1
    how to integ 2x^2 / (1-x^2)
    using partial fracton??
     
  2. jcsd
  3. Jan 3, 2006 #2
    ah....i forgot to type the answer=-2x + ln(1+x) - ln(1-x)
     
  4. Jan 3, 2006 #3

    HallsofIvy

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    First divide to get a "mixed" fraction: [itex]\frac{2x^}{1- x^2}= -2- \frac{2}{1-x^2}[/itex]. Now use partial fractions on the second term.
     
  5. Jan 5, 2006 #4
    Since the numerator and denominator are of the same order, you can turn it into a constant plus a fraction like so

    [tex]\frac{2x^{2}}{1-x^{2}} = \frac{2x^{2}-2+2}{1-x^{2}} = \frac{-2(1-x^{2})}{1-x^{2}} +\frac{2}{1-x^{2}} = -2 +\frac{2}{1-x^{2}}[/tex]

    Now use partial fractions on the second term to get your answer.
     
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