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Answer to \ 2/(x-2)^3 dx

  1. Jan 14, 2007 #1
    1. The problem statement, all variables and given/known data
    \ 2/(x-2)^3 dx
    Basically integrating a perfect cube in the denominator with a constant in the numerator


    2. Relevant equations



    3. The attempt at a solution
    i thought it would be a form of ln(x), but then, that would mean having atleast some x terms in the numerator which are not there, so, how do i do this? Is there a known pre-fixed solution for these things? Like exp(something)?
     
  2. jcsd
  3. Jan 14, 2007 #2
    Use [tex]\int x^ndx = \frac{x^{n+1}}{n+1} + c[/tex]
     
    Last edited: Jan 15, 2007
  4. Jan 14, 2007 #3

    cristo

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    Try writing as 2(x-2)-3. Do you know how to integrate x-3?
     
  5. Jan 14, 2007 #4
    Thanks Arun and Cristo.
    Cristo, I dont know how to integrate x^-3. (i think i might be hopeless, right?)
    Arun, what is the expansion formula you just gave me called? Is there a name for solving by that method? It is applicable to negative powers as well?
     
  6. Jan 14, 2007 #5

    cristo

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    Arunbg's formula is the formula for integrating a polynomial function of x. Thus, to integrate x-3 you would use that formula.
     
  7. Jan 14, 2007 #6
    you want to integrate 2(x-2)-3

    using this [tex]\int x^n = \frac{x^{n+1}}{n+1} + c[/tex]

    the first thing you notice is that you add 1 to the power, then you divided by the new power
    after that you multiply by the differntial of what is inside the brackets
     
  8. Jan 14, 2007 #7
    in your question the power is -3
     
  9. Jan 14, 2007 #8

    cristo

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    The last bit should read: you divide by the derivative of the term inside the brackets.

    (I know it's probably a typo, and it doesn't matter in this case; but it may confuse the OP in future if left uncorrected)
     
  10. Jan 14, 2007 #9
    yes definitely a typo :blushing:
    you always divid by the differential of the inside of the brackets when integrating!
     
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