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Integration by Partial Fractions

  1. Jan 6, 2005 #1
    I'm making a small mistake somewhere, but I can't seem to find it.

    [tex]\int\frac{dx}{(x-1)(1-2x)}[/tex]

    taking the partial fractions

    [tex]1=A(1-2x)+B(x-1)[/tex]

    [tex]A=-1, B=-2[/tex]

    [tex]\int\frac{-1}{x-1} dx+\int\frac{-2}{1-2x}dx[/tex]

    Integrating by substitution, this is what I'm getting

    [tex]-ln(x-1)+ln(1-2x)+C[/tex]

    The correct answer is

    [tex]-ln(x-1)+ln(2x-1)+C[/tex]

    I think I'm making some kind of algebraic sign mistake when taking the second integral, but I just can't find it. I'll go through my solution step by step here and if someone could point out the mistake I would appreciate it.

    [tex]\int\frac{-2}{1-2x}dx[/tex]

    [tex]-2\int\frac{dx}{(1-2x)}[/tex]

    [tex]u=1-2x[/tex] [tex]\frac{du}{dx}=-2[/tex]

    [tex]-2\int\frac{dx}{-2u}[/tex]

    [tex]\int\frac{dx}{u}[/tex]

    [tex]ln(u)+C[/tex]

    [tex]ln(1-2x)+C[/tex]

    Somewhere along the way, I'm missing the spot where I'm supposed to factor out a -1 from the denominator, but where?
     
  2. jcsd
  3. Jan 6, 2005 #2

    arildno

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    Your mistake is rather subtle, it is basically that you think the anti-derivative of 1/x is ln(x); the right expression is ln(|x|) (that is, an absolute value sign must be included)

    To give an example:
    Look at 1/(x-1).
    On the face of it, we would conclude that the antiderivative is ln(x-1).

    But, you could always write:
    1/(x-1)=(-1)/(1-x) (Agreed?)
    But if you integrate the right-hand side, you'll seemingly end up with ln(1-x)

    As you can see, all this makes sense, only if you realize that the correct anti-derivative is ln(|x-1|)
     
    Last edited: Jan 6, 2005
  4. Jan 6, 2005 #3
    Thanks for the quick reply, that explains why it wasn't obvious.
     
  5. Jan 6, 2005 #4
    One more thing.

    Just so I can make sure I understand it. That means I would be free to use either solution?

    The reason I ask is, this integral was part of a larger diff eq problem where I needed ln(2x-1) to verify the solution of a diff eq.

    My TI-89 gives the solution ln(|2x-1|), I can't believe that I didn't notice the || signs the first time I looked at it :yuck:. Is there a reason that ln|2x-1| is more correct then ln|1-2x|?
     
  6. Jan 6, 2005 #5
    ln|2x-1| is identical in every way to ln|1-2x|, I see no reason why you wouldn't be able to use either expression.

    It just so happens I made the exact same mistake a couple of days, and spent hours trying to figure out why Mathematica gave me an answer of ln|(x-1)/(x+1)| while I was getting ln|(1-x)/(1+x)|....
     
  7. Jan 6, 2005 #6

    arildno

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    Is there a reason that ln|2x-1| is more correct then ln|1-2x|?

    Absolutely NONE whatsoever, since we always have: |a|=|-a| for every choice of a.

    So, as long as you keep the absolute value sign in there, you're free to use either expression (since they are equal)
     
  8. Jan 6, 2005 #7
    Great, thanks guys.
     
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