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Homework Help: Integrating by partial fractions

  1. Jun 26, 2006 #1
    Hi, I'm having quite a bit of trouble with this topic. Here's one of the first problems, I dont really understand the method in the book, if someone could show me an easy route, it would help.

    [tex]\int_{0}^{1} \frac {2x+3}{(x+1)^2}dx[/tex]

    Thanks
     
  2. jcsd
  3. Jun 26, 2006 #2

    0rthodontist

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    It's faster not to use partial fractions in this case. You can let u = x + 1.

    What don't you understand about the method in the book?
     
  4. Jun 26, 2006 #3

    arildno

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    Notice that your integrand may be rewritten as:
    [tex]\frac{2x+3}{(x+1)^{2}}=\frac{2x+2}{(x+1)^{2}}+\frac{1}{(x+1)^{2}}=\frac{2}{(x+1)}+\frac{1}{(x+1)^{2}}[/tex]
     
  5. Jun 26, 2006 #4
    I just find that the examples dont really tell me that much, I understand what they do in the examples but then when I get to a question, I rarely never know how to start it.

    arildno, is there a certain way that you came up with that, or do you just see it?
     
  6. Jun 26, 2006 #5

    arildno

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    The certain way is 3=2+1
    And yes, I "just see it". (As you should as well)
     
  7. Jun 26, 2006 #6
    ok, so for this one:

    [tex]\int \frac{x^2}{(x-3)(x+2)^2}dx[/tex]

    would I start out like this:

    [tex]\frac{A}{x-3} + \frac{B}{x+2} + \frac{C}{(x+2)^2}[/tex]

    and then they would just be 1, x, x
     
  8. Jun 26, 2006 #7

    arildno

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    Eeh???
    What do you mean by 1,x,x??
     
  9. Jun 26, 2006 #8

    0rthodontist

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    A, B, and C must be constants (they can't be x). And you know one further thing: The sum you have there must be equal to the original fraction, i.e.
    [tex]\frac{A}{x-3} + \frac{B}{x+2} + \frac{C}{(x+2)^2} = \frac{x^2}{(x-3)(x+2)^2}[/tex]

    Multiply it out, and then you know something else: for this to be true for ALL x the constant terms must be equal on both sides, and so must the coefficients of x, and so must the coefficients of x^2. You will then have three linear equations to solve--namely those that equate the coefficients of each power of x.

    An alternative way that can help you get A, B, and C more rapidly is that you can pick x to be anything you want in that equation, since the equation must be true for all x. So what happens if you multiply both sides by (x - 3) and then set x = 3?
     
    Last edited: Jun 26, 2006
  10. Jun 26, 2006 #9
    ok, i think im getting that part..I have another problem now, im not sure if im going about it the right way.

    [tex]\int_{0}^{1} \frac{x}{x^2 +4x +13}dx[/tex]

    =[tex]\int_{0}^{1}1+[\frac{-x^2-3x-13}{x^2 +4x +13}]dx[/tex]..
    in the book there is an example somewhat like this..they complete the square of the bottom..thats what i did and then substituted..i ended up with this so far:

    [tex]x + \int \frac{-u^2+u-11}{u^2+9}du[/tex]

    if this is right so far, do i just split this up into several fractions and then integrate, or is there more?
     
  11. Jun 26, 2006 #10

    0rthodontist

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    I am not familiar with the technique your book uses. Why did you split off the x? With u = x + 2, you could just turn your original integral into
    [tex]\int_{2}^{3} \frac{u-2}{u^2 + 9}dx[/tex]
    which isn't too hard to integrate.
     
  12. Jun 26, 2006 #11

    arildno

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    Ortho: Not dx, but du..
     
  13. Jun 26, 2006 #12
    so you can do that?its that easy?..why from 3 to 2?
     
  14. Jun 26, 2006 #13
    this isnt the example..this is a question that resembles the example..this is the example..[tex]\int \frac{4x^2-3x+2}{4x^2-4x+3}[/tex]
     
  15. Jun 26, 2006 #14

    0rthodontist

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    Oh yeah, missed that.

    suspenc3, it's from 2 to 3 because when you change from x to u(x), you change the limits from (a to b) to (u(a) to u(b)).
     
  16. Jun 27, 2006 #15

    HallsofIvy

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    The original problem was
    [tex]\int_{0}^{1} \frac{x}{x^2 +4x +13}dx[/tex]
    Completing the square, x2+ 4x+ 13= (x+ 2)2+ 9. If you let u= x+ 2, then x= u- 2 so the integral becomes
    [tex]\int_2^3\frac{u-2}{u^2+1}du[/tex]

    To integrate that, separate it as
    [tex]\int_2^3 \frac{u du}{u^2+1}+ \int_2^3\frac{du}{u^2+1}[/tex]
    The first is an easy v= u2+ 1 substitution and the second is an arctangent.

    I don't see any reason to take out that "1" in
    [tex]\int_{0}^{1}1+[\frac{-x^2-3x-13}{x^2 +4x +13}]dx[/tex]

    I wonder if the example you looked at didn't have a higher power of x in the numerator equal to that in the denominator, say
    [tex]\int_{0}^{1} \frac{x^2}{x^2 +4x +13}dx[/tex].
    Since to use partial fractions you have to have the denominator with degree larger than the numerator, the first thing you would do with that is divide to get 1+ ...
     
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