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Integration by Parts

  1. Oct 24, 2006 #1
    Problem:
    [tex]\int\frac{dx}{a^2-x^2}[/tex]
    My Work:
    [tex]\frac{1}{a^2-x^2}[/tex]
    [tex]=\frac{1}{(a+x)(a-x)}=\frac{A}{a+x}+\frac{B}{a-x}}[/tex]
    [tex]1=A(a-x)+B(a+x)[/tex]
    If x=a, then [tex]1=2Ba[/tex] so [tex]B=\frac{1}{2a}[/tex]
    Thus [tex]1=A(a-x)+\frac{1}{2a}(a+x)[/tex]
    if x=0, then [tex]1=Aa+\frac{1}{2}[/tex] so [tex] A=\frac{1}{2a}[/tex]
    SO
    [tex]\int\frac{dx}{a^2-x^2}[/tex]
    [tex]=\int\frac{\frac{1}{2a}}{a+x}+\frac{\frac{1}{2a}}{a-x}dx[/tex]
    [tex]=\frac{1}{2a}\int\frac{dx}{a+x}+\frac{1}{2a}\int\frac{dx}{a-x}[/tex]
    [tex]=\frac{1}{2a}\ln\mid a+x\mid +\frac{1}{2a}\ln\mid a-x\mid(-1)[/tex]
    [tex]=\frac{1}{2a}(\ln\mid a+x\mid -\ln\mid a-x\mid)[/tex]
    [tex]=\frac{1}{2a}\ln\mid\frac{a+x}{a-x}\mid +C[/tex]

    Table of integrals gives correct answer as
    [tex]=\frac{1}{2a}\ln\mid\frac{x+a}{x-a}\mid +C[/tex]

    My gut feeling is that I messed up integrating [tex]\int\frac{dx}{a-x}[/tex] but I can't find my error.
    Any help would be appreciated.
     
  2. jcsd
  3. Oct 24, 2006 #2
    Its the same thing, because its in absolute values.
     
  4. Oct 24, 2006 #3
    :redface: Thanks, Courtrigrad.
     
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