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Integrating Special Functions

  1. Jul 11, 2012 #1
    This is a problem that came to me when i was doing implicit differentiation and i got curious as to how to integrate a problem like this. I was fascinated by the simplicity if an equation would have a complex integration problem.
    1. The problem statement, all variables and given/known data

    ∫x^x(ln x + 1)dx, Question 1

    ∫x^x dx, Question 2

    2. Relevant equations

    In question 1 the original equation was an innocent looking harmless equation y=x^x.

    In question 2 is what i would have obtain if i have done the following with question 1

    ∫x^x(ln x + 1)dx= ∫x^x . ln x dx + ∫x^x dx [Simply expanded the expression]

    So as it seems expanding the equation does not help me at all.

    3. The attempt at a solution
    Even by substitution or using e^x properties does not help that is

    x= e(ln x)
    x^x= e^(ln x^x)
     
  2. jcsd
  3. Jul 11, 2012 #2

    HallsofIvy

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    Do you have any reason to believe this function can be integrated in terms of elementary functions?
     
  4. Jul 11, 2012 #3
    Well the thing is, this was the result of an implicit differentiation, and since your referring that this cannot be integrated via elementary operations is there any special operations?
     
  5. Jul 11, 2012 #4

    Curious3141

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    Homework Helper

    In Question 1, the substitution [itex]u = x^x[/itex] would immediately solve the integral.

    Expanding it and trying to integrate the individual integrals wouldn't help. But all that proves is that you can't use any technique to calculate a given integral, even if an elementary integral should exist. Sometimes substitution helps (and even then, you have to find the right sub). Sometimes integration by parts helps. There's no reason to suppose that one technique would work in all cases.

    [itex]\int x^x dx[/itex] cannot be expressed in terms of elementary functions. This is not uncommon, most functions don't have integrals expressible in terms of elementary functions.
     
  6. Jul 11, 2012 #5
    Could you kind enough to please show me the steps when you substitute because i am still learning this because my highschool teacher couldnt do this
     
  7. Jul 11, 2012 #6

    Curious3141

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    [tex]\int x^x(1 + \ln x) dx[/tex]

    Substitute [itex]u = x^x[/itex].

    [tex]u = {(e^{\ln x})}^x = e^{x\ln x}[/tex]
    [tex]\frac{du}{dx} = [(x)(\frac{1}{x}) + \ln x].e^{x\ln x} = x^x(1 + \ln x)[/tex]
    [tex]du = x^x(1 + \ln x) dx \Rightarrow dx = {(x^x(1 + \ln x))}^{-1} du[/tex]

    Hence [itex]\int x^x(1 + \ln x) dx = \int x^x(1 + \ln x){(x^x(1 + \ln x))}^{-1} du = \int du = u + C = x^x + C[/itex].

    You can also simply recognise that the derivative of [itex]x^x[/itex] is [itex]x^x(1 + \ln x)[/itex], which means the antiderivative of your integrand is [itex]x^x[/itex].
     
    Last edited: Jul 11, 2012
  8. Jul 11, 2012 #7
    Thank-you very much Curious3141 now i'll show this to my teacher after school holidays so he can put it in my next SAC and i am dead sure no one in my cohort knows how to do it
     
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