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Finding the output of a derivative using integration by partial fractions

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data

    If f is a quadratic function such that f(0) = 1 and
    [tex] \int \frac{f(x)}{x^2(x+1)^3}dx [/tex]
    is a rational function, find the value of f '(0).



    2. Relevant equations



    3. The attempt at a solution

    This question is presented in the context of learning about integration by partial fractions.

    For the quadratic bit I have

    [tex] f(x) = ax^2 + bx + c [/tex] [tex] f(0) = a(0)^2 + b(0) + c = c = 1 [/tex][tex] f'(x) = 2ax + b [/tex] [tex] f'(0) = 2a(0) + b = b [/tex]

    So, I need to find the value of b? Plugging the integral into Wolfram|Alpha gives me

    [tex] \int \frac{ax^2+bx+c}{x^2(x+1)^3}dx = -\frac{a-b+c}{2(x+1)^2} + \frac{b-2c}{x+1} + (b-3c)ln(x) - (b-3c)ln(x+1) - \frac{c}{x} + K [/tex]

    I'll write out the work for the integral by hand when I submit the assignment. For now, I'm not sure how to relate f '(0) = b with any of this to find b, if it is b that I need to find. Thanks for commenting!
     
  2. jcsd
  3. Feb 28, 2012 #2

    Dick

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    If the integral is going to be a rational function, then it shouldn't have any logs in it. What condition will make that true?
     
  4. Feb 29, 2012 #3
    I don't know.

    I messed around with this a bit more and here's what I have.

    [tex] \frac{ax^2+bx+c}{x^2(x+1)^3} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{(x+1)} + \frac{D}{(x+1)^2} + \frac{E}{(x+1)^3} [/tex]

    [tex] ax^2 + bx + c = A(x)(x+1)^3 + B(x+1)^3 + C(x^2)(x+1)^2 + D(x^2)(x+1) + E(x^2) [/tex]

    From there, using the method of partial fractions, you would expand, group, and factor, right? Then setup a system of equations?

    And here's what I have just integrating with the unknown coefficients.

    [tex] \int \frac{ax^2+bx+c}{x^2(x+1)^3} = Aln|x| - \frac{B}{x} + Cln|x+1| - \frac{D}{x+1} - \frac{E}{2(x+1)^2} [/tex]
     
  5. Feb 29, 2012 #4
    Another thought. When x = 0 in the quadratic,

    [tex] a(0)^2 + b(0) + c = A(0)(0+1)^3 + B(0+1)^3 + C(0^2)(0+1)^2 + D(0^2)(0+1) + E(0^2) [/tex]

    simplifies to

    [tex] c = B = 1 [/tex]

    I'm not sure what that tells me. I'll think about it.
     
  6. Feb 29, 2012 #5
    Alright I thought about this again. For the integral to be a rational function, the coefficients on the logs need to equal zero.

    so

    [tex] \int \frac{ax^2+bx+c}{x^2(x+1)^3} = A\cdot ln|x| - \frac{B}{x} + C\cdot ln|x+1| - \frac{D}{x+1} - \frac{E}{2(x+1)^2} [/tex]

    simplifies to, given B = 1 above,

    [tex] = -\frac{1}{x} - \frac{D}{x+1} - \frac{E}{2(x+1)^2}[/tex].

    But I don't think that bit in particular gets me anywhere. However, I can use the fact that A = 0, B = 1, and C = 0 to rewrite

    [tex] ax^2 + bx + c = A(x)(x+1)^3 + B(x+1)^3 + C(x^2)(x+1)^2 + D(x^2)(x+1) + E(x^2) [/tex]

    as

    [tex] ax^2 + bx + c = (x+1)^3 + D(x^2)(x+1) + Ex^2 [/tex].

    Now I can expand, group, and factor the right-hand side of the equation to get the following (if I did my work correctly)

    [tex] 0x^3 + ax^2 + bx + c = Dx^3 + x^3 + Dx^2 + Ex^2 + 3x^2 + 3x + 1 [/tex] [tex] 0x^3 + ax^2 + bx + c = (D + 1)x^3 + (D + E + 3)x^2 + 3x + 1 [/tex]


    At this point can I say that b = 3, which is the answer to the problem statement?
     
  7. Feb 29, 2012 #6

    Dick

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    Yes, f'(0)=b=3. But you didn't need to go through all of that. Look at the integral in your original post. If the log parts are going to vanish you need to have (b-3c)=0. Since you already figured out that c=1. That tells you b=3.
     
  8. Feb 29, 2012 #7
    Haha, well, it was fun to do it the long way too! Thanks for helping Dick!
     
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