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Indefinite Integral Formula

  1. Jan 31, 2007 #1
    There is a simple formula for calculating [tex] \frac{df(x)}{dx} u^n [/tex] where u is a function of x and n is a positive rational number: [tex] \frac{df(x)}{dx} u^n = nu^{n-1} \ast \frac{du}{dx} [/tex]. Is there a similar formula for calculating [tex] \int u^n dx [/tex] where u is a function of x and n is a positive rational number? It would be extremely helpful if there was.

    P.S. I realize that the formula for [tex] \frac{df(x)}{dx} u^n [/tex] can be derived using the chain rule, so I was wondering if maybe the chain rule can somehow be applied in reverse for this problem?
    Last edited: Jan 31, 2007
  2. jcsd
  3. Feb 1, 2007 #2


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    Actually that "f(x)" should not be there. In the Leibniz notation the differentiation operator is simply [itex] \frac{d}{dx} [/itex].

    As for the question itself, there's no general method for computing [itex] \int u^{n}(x) {} dx [/itex] for arbitrary "u(x)". In most cases, one can't find the antiderivative in terms of elemetary functions, even though "u(x)" may be elementary.
  4. Feb 1, 2007 #3


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    The "reverse" of the chain rule is basically integration by substitution. The problem is that when differentiating, using the chain rule, we are free to multiply by du/dx no matter what it is. When integrating, using substitution, that du/dx must already be in the integral- unless it is a constant, we can't just "put it in".

    [tex]\int u^n \frac{du}{dx}dx= \int u^n du= \frac{1}{n+1}u^{n+1}[/tex]
    But that "du/dx" must already be in the integral.

    The one exception to that is, as I said, when du/dx is a constant- when u is linear.
    [tex]\int (ax+b)^n dx= \frac{1}{a(n+1)}(ax+b)^{n+1}[/tex]
    By letting u= ax+ b so that du= a dx and dx= (1/a)du.
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