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Homework Help: Integration Question

  1. May 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Q. Solve ∫ (5x^2 - 42x + 24)/(x^3 - 10x^2 + 12x + 72) dx

    2. Relevant equations
    ∫ x^n dx = ∫ (1/n+1)x^(n+1) + C
    ∫ 1/x dx = ln x + C

    3. The attempt at a solution

    Hi everyone,

    Here's what I've done so far:

    (i) I factored the bottom equation into (x + 2)(x - 6)^2

    and then separated the fraction 1/[(x + 2)(x - 6)^2] into a sum of three fractions of the form:
    A/(x + 2) + B/(x - 6) + C/[ (x-6)^2 ]

    and found that A = 1/64, B = -1/64, C = 1/8

    (ii) The only way I can see to go from here is to split the top integral into three integrals:

    (1/64) ∫ (5x^2 - 42x + 24)/(x + 2) dx + (-1/64) ∫ (5x^2 - 42x + 24)/(x - 6) dx + (1/8) ∫ (5x^2 - 42x + 24)/(x-6)^2 dx

    and solve them all individually using substitution, letting u = x+2, x-6 and x-6 respectively.

    The answer I get in this way is: (5/8)x - 21/4 +2ln(x+2) + 3ln(x-6) + 6/(x-6) + C, C = constant of integration,

    which is incorrect, according to the WebWork site, but I can't find any algebraic errors.
    So, I'm thinking perhaps what I do in step (ii) is not the best way to do it.

    Has anyone any suggestions?

    Thanks for any help!
  2. jcsd
  3. May 9, 2009 #2


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    Science Advisor
    Homework Helper

    You don't want to do it that way. You want to split (5x^2 - 42x + 24)/((x+2)*(x-6)^2) into three fractions of the form A/(x + 2) + B/(x - 6) + C/[ (x-6)^2 ] with A, B and C constant.
    Last edited: May 9, 2009
  4. May 9, 2009 #3


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    Homework Helper

    You have split your fractions correctly. However splitting [itex]1/((x-6)^2(x+2))[/itex] instead of the entire integrand, like Dick suggests, results in you having to integrate three integrals which are very sensitive to mistakes. I checked the expression you wrote down in (ii) in mathematica and it does give you the correct answer so you must have made a mistake integrating somewhere. However I suggest you do it Dick's way, it gives easy values for A, B and C (integers) and results in three very easy integrals.
    Last edited: May 9, 2009
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