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

  • Thread starter supercali
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  • #1
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Homework Statement



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

Homework Equations


---

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]
 

Answers and Replies

  • #2
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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]
And how did you get to this step?
 
  • #3
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[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]
 
  • #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.
 
  • #5
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thats no a polynomial how do i use partial fractions here?
 
  • #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.
 
  • #7
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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
 
  • #8

Homework Statement



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

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.
 
  • #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
 
  • #10
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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:
  • #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 :(
 
  • #12
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actually i got the answer with wolfram mathematica so i dont think its the program and in any case never trust a computer
 
  • #13
Gib Z
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  • #14
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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
 
  • #15
Gib Z
Homework Helper
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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
 
  • #16
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happy birthday
hope you will do better next time
 
  • #17
1,752
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Sigh. That wasn't hard at all, Rainbow Child wins!
 

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