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Calc 2 Limits

  1. Nov 15, 2007 #1
    1. The problem statement, all variables and given/known data

    Integral of (3x+2)/x(x+2)^2+16x

    2. Relevant equations

    3. The attempt at a solution

    That breaks down to

    A/x + Bx+c/x^2+4x+20

    so 3x+2 = Ax^2+4x+20 + Bx^2 + Cx

    then I found the values of A b and C then I cant figure out what to do please help
  2. jcsd
  3. Nov 15, 2007 #2
    What did you get for values of a,b,c?

    And by the way, is that supposed to be

    [tex] \frac{3x+2}{x(x+2)^2+16x} [/tex]?

    because the way it's written is

    [tex] \frac{3x+2}{x} (x+2)^2 + 16x [/tex]
  4. Nov 15, 2007 #3
    yea thats what its suposed to be
  5. Nov 15, 2007 #4
    And did you by chance get A= 1/10, B = -1/10, C= 26/10?
  6. Nov 15, 2007 #5
  7. Nov 15, 2007 #6
    I dont know what to do after this
  8. Nov 15, 2007 #7
    well, the [itex]\frac{1}{x} [/itex] is pretty easy to handle right? So we'll just focus on the other term. Now, in this case it's better if we express [itex] x^2+4x+20 [/itex] as [itex] (x+2)^2+16[/itex].

    Make the substitution [itex] x+2 = 4 \tan(\theta) [/itex] and don't forget that in this case [itex] dx = 4 \sec^2(\theta) d\theta [/tex]. Substitute everything into your integral and see if it simplifies a bit.
  9. Nov 15, 2007 #8
    I dont know what to do after this
  10. Nov 15, 2007 #9
    If you do what I've said to do, and you do it correctly, your integral will become much easier, so just stick with it.
  11. Nov 15, 2007 #10
    I dont know what to do after this
  12. Nov 15, 2007 #11
    Well why don't you show me what you've got so far and we'll see if we can't see where the problem is, because if you done it correctly the integral is blatantly obvious.
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