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Homework Help: Integration by substitution

  1. Jul 10, 2010 #1
    Hi there,

    I am having difficulty with one aspect of intergration by substitution where the substituion of a square root is U^2, wondering if anyone can help.

    Problem:

    Integral of: 2x√(3x-4) dx by substituting U^2 = 3x-4

    Would du^2/dx = 3 therefore 1/3 du^2 = dx (I think this is where im going wrong)


    Im coming out with an answer of: 2/45(9x+8)(3x-4)^3/2 + k

    However the answer in the book is: 4/135(9x+8)(3x-4)^3/2 + k

    Any help would be greatly appreciated.

    Regards

    Dave
     
  2. jcsd
  3. Jul 10, 2010 #2
    Your error is that the derivative of u^2 isn't du^2.

    [tex]u^{2}=3x-4[/tex]

    [tex]2u du=3dx[/tex]

    [tex]du=\frac{3}{2\sqrt{3x-4}}dx[/tex]
     
    Last edited: Jul 10, 2010
  4. Jul 10, 2010 #3
    Ah brilliant! Thanks for the help mate!
     
  5. Jul 10, 2010 #4

    Mark44

    Staff: Mentor

    And really, the substitution is u = sqrt(3x - 4). If u >= 0, this is equivalent to u^2 = 3x - 4.
     
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