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Indefinite Integrals & Substitution Rule

  1. Apr 24, 2010 #1
    1. The problem statement, all variables and given/known data

    2. The attempt at a solution

    Last edited: Apr 25, 2010
  2. jcsd
  3. Apr 24, 2010 #2


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    It's not even close. Since when does the integral of du/sqrt(u) have an ln in it? And shouldn't you be using du=(-5)dx to replace dx with du?
  4. Apr 24, 2010 #3
    I forgot about the square root.
    how about this?

    I= ∫ dx/sqrt(2−5x)
    = ∫ [1/sqrt(2−5x)]dx
    u= 2-5x
    du= -5dx
    (-1/5)du = dx

    ∫ (-1/5)du/sqrt(u)
    (-1/5) ∫ du/u^(1/2)
    (-1/5) 1/ (2/3)u^(3/2)
    - 1/[(10/3)u^(3/2)]
    - 1/[(10/3)(2-5x)^(3/2)]
    Last edited: Apr 24, 2010
  5. Apr 24, 2010 #4


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    du/sqrt(u) is u^(-1/2)du. You've still got the power rule wrong.
  6. Apr 25, 2010 #5


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    Do remember there is a way to check your work -- what relationship is a function supposed to have with its anti-derivatives?
  7. Apr 25, 2010 #6
    Yeah, I already know that.

    I don't trust my math, that's why I want to know if I'm doing this correctly.
    Last edited: Apr 25, 2010
  8. Apr 25, 2010 #7
    ok nevermind I got it.....
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