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Am I integrating this right: (x^2 + 3x + 11)/(x+2)^4 ?

  1. Mar 3, 2013 #1
    Am I integrating this right: (x^2 + 3x + 11)/(x+2)^4 ???

    1. ∫(x2 + 3x + 4)/(x+2)4dx


    2. With these sorts of problems, I think about integration by partial fractions.

    So in this case, the denominator factors are all the same, so I have to make each one with different exponents.
    I wrote:

    A/(x + 2) + B/(x + 2)2 + C/(x + 2)3 + D/(x + 2)4

    I multiplied to get common denominators:

    A(x + 2)3 + B(x + 2)2 + C(x + 2) + D = x2 + 3x + 11

    Now I need to figure out what A, B, C, and D equal.

    When I substitute x=-2, D=9

    A(x + 2)3 + B(x + 2)2 + C(x + 2) + 9 = x3 + 3x + 11

    I think I am supposed to take the derivative now to figure out the other values.

    2x + 3 = 3A(x + 2)2 + 2B(x + 2) + C

    When I substitute x=-2, C= -1

    Now, am I able to take the derivative again?
    That's what I did next:
    Substituting x=-2,

    2 = 6A(x + 2) + 2B
    B=1

    If x=1,

    2 = 6A(x + 2) + 2(1)
    A=0

    ...Am I even doing this right? :/
    I don't want to go on if I'm not even doing this part right!
    Thank you very much! :D
     
  2. jcsd
  3. Mar 3, 2013 #2

    Dick

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    If your original integrand was (x^2+3x+11)/(x+2)^4 (you probably have a typo), then, yes, you got the partial fractions right.
     
  4. Mar 3, 2013 #3
    Oh, yes, was a typo! Thanks for letting me know it's right so far! :)
     
  5. Mar 4, 2013 #4
    Alright, now I have:

    ∫(0/x+2) + (1/(x+2)2) - 1/(x+2)3 + 9/(x+2)4dx

    U-substitution!

    u= x+2
    du=dx

    ∫u-2 - u-3 + 9u-4du

    = -u-1 - u-2/-2 + 9u-3/-3

    = -1/(x+2) + 1/(2(x+2)2) - 2/(x+2)-3

    ...Is this right?
    Thanks so much! :)
     
  6. Mar 4, 2013 #5

    SammyS

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    You can check by taking the derivative of the result.

    By the way: This problem can be done more easily by using u-substitution at the outset, with u = x+2 .
     
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