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U-substitution - Where does the dx go?

  1. Mar 9, 2010 #1
    I'm having serious trouble with the concept of u-substitution. Using this as an example
    int 2x(x2−1)4
    I make u = x2−1
    first thing I don't get is why du/dx = 2x is rearranged to du = 2xdx. Second thing I don't get is where the dx dissappears to. In this method is 2xdx just being represented as du? The part that confuses me is how you can represent something in an integrand as the derivative of another term in the integrand. I'm completely lost there.
  2. jcsd
  3. Mar 9, 2010 #2
    It looks better if you write this out:

    [tex] \int f(g(x))g'(x)dx = \int f(u) du [/tex]

    this is the substitution formula, the [tex] dx [/tex] part is adjusted by [tex] g'(x)dx [/tex]
    to act as [tex] du [/tex]

    so, 2x is the g'(x) part, x^4 is your f(x) part, and g(x) = x^2 -1
  4. Mar 9, 2010 #3
    Ah that explains it. I wasn't thinking of the dx being replaced by du. Thanks alot. Its funny how something which seemed incomprehendable falls into place in a matter of seconds once after making a small connection. That LaTeX code is fairly cool.
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