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Unsolvable Problem?

  1. Apr 30, 2008 #1
    [SOLVED] Unsolvable Problem?


    How can we find the antiderivative of [tex]2(1 + lnx)(x^x)^2[/tex]

    As it stands it is not algebraically solvable.
    This question is from my calculus' book, but the solutions are not provided...

    Does anyone know of any mthod we can use to integrate this?

  2. jcsd
  3. Apr 30, 2008 #2
    Substitution. [tex]u=(x^x)^2[/tex].
  4. Apr 30, 2008 #3


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    Oh, that is clever!
  5. Apr 30, 2008 #4
    actually that is the antiderivative x^(2x)
  6. Apr 30, 2008 #5

    [tex]\int 2(1+\log x)(x^x)^2\, dx[/tex]: [tex]u=(x^x)^2[/tex]; by logarithmic differentiation we find [tex]du=2(x^x)^2(\log x + 1)[/tex] so the integral is [tex]\int\, du = u = (x^x)^2 = x^{2x}[/tex].
  7. Apr 30, 2008 #6
    so how did you figure out that substitution?
  8. Apr 30, 2008 #7
    If I had to guess, I'd have to say it was either
    1. a barter for his soul with some demon/deity
    2. divine intervention
    3. ??? can't think of any other possibility

    Seriously, that was clever
  9. Apr 30, 2008 #8


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    I suspect that he realized that, since the derivative of xx involves a logarithm, and there was a logarithm in the problem, he should try something like u= xx, tried it, saw that it didn't quite work and then played with it until he found something that did.

    Two many math students seem to think that the way you solve a math problem is to stare at the paper until the solution springs, like Venus, full grown from your forehead.
  10. Apr 30, 2008 #9
    maybe but writing simply for the sake of writing isn't effective either.
  11. Apr 30, 2008 #10
    I dunno, I just saw it and tried it and voilà, it worked.

    I knew the answer was [tex](x^x)^2[/tex] before I solved the problem, which may or may not have helped, but probably did.
  12. Apr 30, 2008 #11
    Also I left out a [tex]dx[/tex] in my first post.
  13. Apr 30, 2008 #12
    Also (and sorry for the triple post!) [tex]u=x^x[/tex] would have worked just as well.
  14. Apr 30, 2008 #13

    Here is my working, by using substitution:

    Let I = ∫2(1+lnx)(x^x)2 dx

    Let u = x^x

    ∴ lnu = lnx^x

    ∴ lnu = xlnx

    ∴ 1/u × dudx = x×1/x + 1×lnx

    ∴ du/dx = u ×(1+lnx)

    ∴ du = x^x^×(1+lnx)dx

    ∴ dx = du / x^x^×(1+lnx)

    ∴ I = ∫2(1+lnx)(x^x)^2 dx

    = ∫2(1+lnx)(x^x)^2{du / [x^x×(1+lnx)]}

    = ∫2 x^xdu

    = ∫ 2u du

    = u² + c

    = (x^x)^2 + c

    = x^2x + c

    I didn't know if it was correct but now since you have the same solution as me then it must be right...

    Many thanks.
    Last edited: Apr 30, 2008
  15. May 1, 2008 #14
    Yep, that seems right although it was hard to follow somewhat because you didn't use LaTeX...
  16. May 1, 2008 #15


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    That makes for quite an amusing thread title! :tongue2:
  17. May 1, 2008 #16


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    I concur.
  18. May 1, 2008 #17
    Oh man I was just thinking that! You have mah brain, cristo.
  19. May 6, 2008 #18
    my calculus teacher taught us this a couple weeks ago. its pretty useful
  20. May 8, 2008 #19

    I know...about the title... :blushing: :redface:

    But it looked unsolvable to me because the original question looked like this: [tex]2(1 + lnx)x^{x^2}[/tex]

    I reckon they forgot to include the brackets. But my teacher then told me to include the brackets and then solve it that way.
    [tex]2(1 + lnx)(x^x)^2[/tex]

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