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Integrating Rational Function

  1. Jan 31, 2013 #1
    1. The problem statement, all variables and given/known data
    integrate the following:


    2. Relevant equations
    ∫(x/(x-1)^3


    3. The attempt at a solution
    i've tried u-substitution, finding an inverse trig function that matched the formula, and still can't figure out how to solve this problem.

    u-subtitution for u=x gives the same problem. u-subsitution for x-1 gives du =1 which does not match the problem.
     
  2. jcsd
  3. Jan 31, 2013 #2

    Dick

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    u=(x-1) gives du=dx which does match the problem. If you are worried about the x in the numerator, if u=x-1, then x=u+1.
     
  4. Jan 31, 2013 #3
    Its a fairly easy integral..dont go all complicated when you cant find the answer just stay on the ground bro :tongue2: sometimes few problems can be easily solved if you just view it from a different angle

    this should work. you would get u+1/u^3 du which you split and integrate
     
  5. Jan 31, 2013 #4

    Ray Vickson

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    No, you will not get [tex]u + \frac{1}{u^3},[/tex] which is what you wrote! If you really mean [tex] \frac{u+1}{u^3}, [/tex] use parentheses, like this: (u+1)/u^3.
     
  6. Jan 31, 2013 #5

    Dick

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    Good advice but use more parentheses. You could easily mistake u+1/u^3 for u+(1/u^3) when you meant (u+1)/u^3.
     
  7. Jan 31, 2013 #6
    lol thanx for pointing it out..my bad :shy:
     
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