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Integration of a fraction

  1. Oct 27, 2012 #1
    I have trouble solving the following integral [itex]\frac{x-6}{x^2-4}[/itex]

    If I let u= x^2 - 4, I'm left with 1/2du = x dx

    and I'm stuck trying to get rid of that 6 in the numerator?

    If someone can help me out, that would be good.
     
  2. jcsd
  3. Oct 27, 2012 #2

    haruspex

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    Do you know how to rewrite expressions like 1/((ax+b)(cx+d)) as a sum of simpler fractions?
     
  4. Oct 27, 2012 #3
    There's another method besides the one that @haruspex mentioned. You can also rewrite the numerator so that it contains a du.

    For example,
    [itex]\frac{x + 2}{\frac{1}{2}x^2 + 3x}[/itex]

    Letting
    [itex]u = \frac{1}{2}x^2 + 3x[/itex],

    [itex]du = x + 3[/itex].

    Then the original numerator can be manipulated algebraically to get a proper substitution involving du, plus an additional term.

    Sticking to my example,
    [itex]\frac{x + 2}{\frac{1}{2}x^2 + 3x}[/itex]

    [itex]\frac{x + 3 - 1}{\frac{1}{2}x^2 + 3x}[/itex]

    [itex]\frac{(du) - 1}{u}[/itex]
     
  5. Oct 27, 2012 #4

    haruspex

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    You mean [itex]du = (x + 3)dx[/itex]
    Filling in the details exposes a flaw:
    [itex]\frac{x + 2}{\frac{1}{2}x^2 + 3x}dx[/itex]
    [itex]\frac{x + 3 - 1}{\frac{1}{2}x^2 + 3x}dx[/itex]
    [itex]\frac{du - dx}{u}[/itex]
     
  6. Oct 27, 2012 #5

    SammyS

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    Actually, [itex]du = (x + 3)dx[/itex]
    You end up with only being able to use u for part of the integrand.

    [itex]\displaystyle \frac{x + 2}{\frac{1}{2}x^2 + 3x}\quad\to\quad\frac{(du) - dx}{u} \quad\to\quad \frac{du}{u}-\frac{dx}{\frac{1}{2}x^2 + 3x}[/itex]
     
  7. Oct 27, 2012 #6
    Why are you guys doing the part where its du - dx in the numerator I don't get where the dx part is coming from? and whats its value?
     
  8. Oct 27, 2012 #7

    haruspex

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    This part of the thread was triggered by SithsNGiggles' post, which had an error in it. SammyS and I jumped in to stop you being misled by it.
    Please go back to my first post.
     
  9. Oct 27, 2012 #8

    SammyS

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    I was merely responding to SithsNGiggles's post -- that's why I 'Quoted' it. His suggestion was erroneous.

    What haruspex suggested is the way to go. That is, to use partial fraction decomposition to write [itex]\displaystyle \frac{x-6}{x^2-4}[/itex] as the sum of two fractions, one with a denominator of x-2 the other with a denominator of x+2 .
     
    Last edited: Oct 27, 2012
  10. Oct 27, 2012 #9
    ohh is that the integral where we do the A/x-2 + B/x+2 thing? I don't really remember doing that a lot I forgot what the values of A and B would be
     
  11. Oct 27, 2012 #10

    haruspex

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    You don't have to remember. Just write down that your fraction equals A/(x-2) + B/(x+2), multiply out, and find the values of A and B that make all the coefficients match up.
     
  12. Oct 28, 2012 #11
    Oh, sorry about that. I forgot to show the step where the fraction was split to du/u + dx/u. Thanks for catching that.
     
  13. Oct 29, 2012 #12
    ok guys thanks I solved this problem pretty sure I'm correct to,
    i got
    2ln|x+2| + lin|x-2| + C

    hmm the solution says - lin|x-2| on the second part , why is it - rather than +
     
  14. Oct 29, 2012 #13

    Dick

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    What did you get for A and B in the partial fractions expansion? You should find your - sign there.
     
  15. Oct 29, 2012 #14
    I had
    A+B=1
    -2A + 2B= -6

    multiplied top by 2 then substracted with bottom,
    2A + 2B = 2
    -2A +2B = -6

    Subtracting the top with bottom I got
    4A = 8
    A = 2,
    sub A into the first equation

    2 + B = 1
    2-1 = -B

    -1 = B
    OH WHOOPS I didnt make B negative when I moved it to the other side!
     
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