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Find the integral 2^x 3^x / 9^x - 4^x

  1. Jan 10, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\int{\frac{2^x3^x}{9^x-4^x}dx[/tex]

    2. Relevant equations
    ---

    3. The attempt at a solution
    I was able to bring it to this: (though i dont know if it helps or not)

    [tex]\frac{1}{2} \int{\frac{3^x+2^x}{3^x-2^x}-\frac{3^{2x}+2^{2x}}{3^{2x}-2^{2x}}}dx[/tex]
     
  2. jcsd
  3. Jan 10, 2008 #2
    And how did you get to this step?
     
  4. Jan 10, 2008 #3
    [tex]\int{\frac{2^x3^x}{9^x-4^x}dx=\frac{1}{2} \int{\frac{22^x3^x}{3^{2x}-2^{2x}}dx=\frac{1}{2} \int{\frac{23^x2^x+3^{2x}+2^{2x}-3^{2x}-2^{2x}}{3^{2x}-2^{2x}}}dx=\frac{1}{2} \int{\frac{{(3^x+2^x)}^2-3^{2x}-2^{2x}}{(3^{x}-2^{x})(3^{x}+2^{x})}}dx=\frac{1}{2} \int{\frac{3^x+2^x}{3^x-2^x}-\frac{3^{2x}+2^{2x}}{3^{2x}-2^{2x}}}dx[/tex]
     
  5. Jan 11, 2008 #4

    Gib Z

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    If you know double integrals, try the substitutions u= 2^x, v = 3^x. if not, try using partial fractions.
     
  6. Jan 11, 2008 #5
    thats no a polynomial how do i use partial fractions here?
     
  7. Jan 11, 2008 #6

    Gib Z

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    Actually never mind I didn't pay attention to your post 3, where you already did =] Partial fractions doesn't always need polynomials (I'm not sure if this method is still called partial fractions though, since its not completely analogous, but same idea).

    In the end theres no point though, I've tried a lot of methods, then consulted the Integrator; there is no non-trivial solution for the indefinite integral.
     
  8. Jan 11, 2008 #7
    this is strange because this qusteion was in my calculus book right at the begining after the part that they elxplain about all the basic laws of integration
     
  9. Jan 11, 2008 #8
    First divide the numerator and the denominator of the fraction with [itex]3^{2\,x}[/itex] and then make the change of variables

    [tex]u=(\frac{2}{3})^x,\, d\,x=\frac{d\,u}{\ln\frac{2}{3}\,u}[/tex]

    so

    [tex]I=\int\frac{(\frac{2}{3})^x}{1-(\frac{2}{3})^{2\,x}}\,d\,x=\int\frac{u}{1-u^2}\,\frac{d\,u}{\ln\frac{2}{3}\,u}=\frac{1}{\ln\frac{2}{3}}\,\int\frac{1}{1-u^2}\,du[/tex]

    and the intergration is trivial with partial fractions.
     
  10. Jan 11, 2008 #9

    Gib Z

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    Why is the integrator is horrible at integrals now..and me even more so :( I've lost my touch lol
     
  11. Jan 11, 2008 #10
    dont to be too hard on your self i dont think it was an easy one
    and by the way Rainbow Child thank you
     
    Last edited: Jan 11, 2008
  12. Jan 11, 2008 #11

    Gib Z

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    I'm actually much more worried about the integrator's response now though :( I used to go to it as a resource, I thought "If it can't do it, it's not possible". In fact, most people do think that, Mathematicia is a very widely used and trusted mathematical software. It even says it self, if the integrator can't find the integral, it's most likely there is no such formula. And this was a (relatively) basic integral :(

    I should have got it too :(
     
  13. Jan 11, 2008 #12
    actually i got the answer with wolfram mathematica so i dont think its the program and in any case never trust a computer
     
  14. Jan 11, 2008 #13

    Gib Z

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  15. Jan 11, 2008 #14
    even here it gives the right answer
    type this into the integrator:
    ((2^x)*(3^x))/((9^x)-(4^x))
    and any case i have the program and there it gives it more nicely
     
  16. Jan 11, 2008 #15

    Gib Z

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    The first time I just put in;
    (2^x 3^x)/(9^x - 4^x)

    Sigh VERY bad days :( Still tired from yesterday, lots of partying. Was my birthday =D
     
  17. Jan 11, 2008 #16
    happy birthday
    hope you will do better next time
     
  18. Jan 11, 2008 #17
    Sigh. That wasn't hard at all, Rainbow Child wins!
     
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