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Stuck at partial fraction problem.

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data
    Evaluate ∫1/x(x-1)2, by using the partial fraction method


    2. The attempt at a solution

    Stating in partial fraction form:

    1/(x(x-1)2 = A/x + B/(x-1) + C/(x-1)2

    1 = Ax(x-1)2/x + Bx(x-1)2/(x-1) + Cx(x-1)2/(x-1)2

    1 = A(x-1)2 + Bx(x-1) + Cx

    and this is where I am stuck because I must compare the coefficients to 1...any idea how I can go on after this?
     
  2. jcsd
  3. Oct 2, 2011 #2
    you have to express your expression like
    1 = A(x-1)^2 + Bx(x-1) + Cx = Dx^2 + Ex + F

    express D,E,F in terms of A,B and known numbers.

    and you know that (Dx^2 + Ex + F = 1) for any x. To make it always equal to 1, your coefficients near x^2 and x must be 0 and your free coefficient must be 1.
     
  4. Oct 2, 2011 #3

    epenguin

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    Gold Member

    That is exactly what you do. You'd have to expand and collect first to know the coefficients of x0, x1, x2.

    A bit pedestrian. As this is an identity true for all x, what happens when you make x = 1, when you make x = 0 ?
     
  5. Oct 2, 2011 #4
    ok, continuin with the given information from the people above:

    1 = A(x2-2x+1) + Bx2-B + Cx

    1 = Ax2-2Ax+A + Bx2-B + Cx

    Sorting through, then I will get:

    1 = Ax2 + Bx2 -2Ax + Cx + A - B

    1 = (A+B)x2 - (2A-C)x + (A-B)

    From the above equation:

    A + B = 0

    2A - C = 0

    A - B = 1

    Therefore A=1/2 , B= -1/2, C = 1

    Rewriting the partial fraction form:

    ∫1/x(x-1)2 dx = ∫ (1/2)/x + (-1/2)/(x-1) + 1/(x-1)2 dx

    = ∫ x/2 dx + ∫ -(x-1)/2 dx + ∫ 1/(x-1)2 dx

    = x2/4 - (x2/4 - x/2) + ∫1/(x-1)2

    Do I have to re-apply partial fraction at the last part as well?
     
  6. Oct 2, 2011 #5
    You can certainly collect like terms: as a matter of fact that's how I used to do things also. However, epenguin described a really useful device for getting the coefficients in a much faster way.

    Since you are trying to solve 1 = A(x-1)^2 + Bx(x-1) + Cx, this must hold for all x right? Well, try plugging x = 0 into this to find that (-1)^2A =1. Similarly, plugging in x=1 will give you C(1) =1. You can use these to solve for B.

    Your way is still correct, but this way can save you the task of simplifying and avoid computational errors.

    Edit: Just saw that the 2 was changed to a squared. Fixing my post to reflect that.
     
    Last edited: Oct 2, 2011
  7. Oct 2, 2011 #6
    To which part? [itex] \int\frac1{(x-1)^2} dx [/itex]? This can be solved with substitution.

    Edit: Made a mistake with my comment
     
    Last edited: Oct 2, 2011
  8. Oct 2, 2011 #7
    All right, thanks guys. I've also done the substitution part.

    u=x-1
    du=dx

    Finally got it.

    At Kreizhn: Its "1 = A(x-1)^2 + Bx(x-1) + Cx", not "1 = A(x-1)2 + Bx(x-1) + Cx" or maybe you meant something else? :P
     
  9. Oct 2, 2011 #8
    Thanks for point that out. I actually copied and pasted that line, so it must have been changed from somewhere. Anyway, I've fixed that post to reflect that fact. Also note that, as I warned earlier, you made a computational mistake in evaluating your partial fractions. You expanded Bx(x-1) = Bx^2 - B which is not correct. Give the method I mentioned a try and you'll see the answers practically throw themselves at you.
     
  10. Oct 2, 2011 #9
    oh dayum...dayum....d.a.y.u.m....

    Thank Kreizhn for pointing that out, ouch.

    So from the method that you were saying:

    1= A(x-1)2 + Bx(x-1) + Cx

    Substitute x=0 to the equation, I will get:

    1 = A(0-1)2 + B(0)(0-1) + C(0)

    1= A(-1)^2

    A=1

    Substitute x =1 to the equation, I will get:

    1 = A(1-1)2 + B(1)(1-1) + C(1)

    C=1
     
  11. Oct 2, 2011 #10
    Exactly. And you can use any other value to get an equation for B. Try x=2, that works nicely.
     
  12. Oct 2, 2011 #11
    So, substituting x = 2, A = 1 and C = 1, I will get:

    1 = A(2-1)^2 + B(2)(2-1) + C(2)

    1 = A^2 + 2B + 2C

    1= 1^1 + 2B + 2

    2B = -2

    B = -1

    Got it, thx a lot. Much appreciated. Learned something new here. When you say "can use any other value to get an equation for B", can it be implemented to find A and B as well? or I must use x = 0 and x = 1 to initially find A and C?
     
  13. Oct 2, 2011 #12
    You can use anything. It's just that x=0 and x=1 were particularly useful because they happened to "kill" some of the other variables. In general, you won't be able to get rid all but one variable, but this gives a really easy way of getting the system of equations without needed to collect like terms.
     
  14. Oct 2, 2011 #13
    Ok, thanks for the info.
     
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