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tsghernan

  • 2 years ago

I can not solve this integral

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  1. cnknd
    • 2 years ago
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    cool story bro

  2. tsghernan
    • 2 years ago
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    \[\int\limits_{1}^{9} \frac{ 1 }{ \sqrt{x*\sqrt{1+\sqrt{x}}} } \delta x \] (The integral is between 9 and 1

  3. g152xx
    • 2 years ago
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    @TuringTest is here so have no fear ! :)

  4. TuringTest
    • 2 years ago
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    Why do some people think I am a math demi-god? I have no idea.... at least not yet.

  5. cnknd
    • 2 years ago
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    try rewriting this with fractional powers instead of nested sqrts

  6. cnknd
    • 2 years ago
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    then do a u-substitution where u = sqrt(x) (i'm just eyeballing this but it seems like this should work)

  7. TuringTest
    • 2 years ago
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    you ought to try some of the things you suggest or it just becomes a flood of ideas rather than an clear approach that will help the asker.

  8. tcarroll010
    • 2 years ago
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    No, substitute u = 1 + x^(1/2) Then du = (1/2)x^(-1/2) dx and then easy to solve from that point.

  9. cnknd
    • 2 years ago
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    heh, same thing, just off by 1 (du is still the same)

  10. TuringTest
    • 2 years ago
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    @tcarroll010 yes, I agree :)

  11. tcarroll010
    • 2 years ago
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    @cnknd , it is necessary to isolate that term of "1" into u along with the x^(1/2)

  12. cnknd
    • 2 years ago
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    i don't think it's all that difficult to integrate (1+u)^(-1/4) rather than u^(-1/4)

  13. tcarroll010
    • 2 years ago
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    \[2\int\limits_{}^{}\frac{ du }{ \sqrt{u} }\]

  14. tcarroll010
    • 2 years ago
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    \[= 4\sqrt{u}\]Now, just put the expression for u back in.

  15. tcarroll010
    • 2 years ago
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    \[4\sqrt{1 + \sqrt{x}}\]And you're done.

  16. cnknd
    • 2 years ago
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    @tcarroll010 you forgot the a 2nd sqrt. it's \[2\int\limits_{2}^{4}\frac{ du }{ \sqrt{\sqrt{u}} }\]

  17. tcarroll010
    • 2 years ago
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    Did not forget. Go back and look at the du equation. sqrt(x) is already in it.

  18. cnknd
    • 2 years ago
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    original eqn had 3 sqrts

  19. tcarroll010
    • 2 years ago
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    They get resolved in the du equation. Simply take the derivative of my answer and you'll see.

  20. cnknd
    • 2 years ago
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    the first sqrt from the original expression covers both the x and the sqrt(1+sqrt(x)) unless my eyes are deceiving me.

  21. g152xx
    • 2 years ago
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    @cnknd - @tcarroll010 is right, take a look again.

  22. tcarroll010
    • 2 years ago
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    Simply rewrite the original equation to\[\frac{ 1 }{ \sqrt{x}\sqrt{1 + \sqrt{x}} }\]and then it will be clearer. Again, just take the derivative of my answer and you'll get the original equation, written either way.

  23. cnknd
    • 2 years ago
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    original integrand was: \[\frac{ 1 }{ \sqrt{x \sqrt{1+\sqrt{x}}} }\]

  24. tcarroll010
    • 2 years ago
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    yes, which is the same as my re-write.

  25. asnaseer
    • 2 years ago
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    @cnknd - @tcarroll010 is right - he is making use of this:\[u = 1 + x^{1/2}\]Then\[du = (1/2)x^{-1/2} dx=\frac{dx}{2\sqrt{x}}\]

  26. cnknd
    • 2 years ago
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    let \[A = \sqrt{1+\sqrt{x}}\] the original integrand would be: \[\frac{ 1 }{ \sqrt{xA} }\] and your rewrite is: \[\frac{ 1 }{ \sqrt{x}*A }\] im pretty sure those are different

  27. tsghernan
    • 2 years ago
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    I can solve it. Thank you guys

  28. cnknd
    • 2 years ago
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    if you want to double check your answer, here's wolframalpha's solution: http://www.wolframalpha.com/input/?i=integrate+%281%2Fsqrt%28x*sqrt%281%2Bsqrt%28x%29%29%29%29

  29. tsghernan
    • 2 years ago
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    tcarroll010 was right. The best way is substitute u = 1 + x^(1/2)

  30. asnaseer
    • 2 years ago
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    oh - sorry @cnknd - I thought you were wondering where the \(\sqrt{x}\) went. you are right - there should be an extra sqrt there.

  31. asnaseer
    • 2 years ago
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    @tsghernan - please take note of what @cnknd was saying as well - there is an extra sqrt to put in

  32. cnknd
    • 2 years ago
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    thank you sir moderator :) and yea u = 1 + x^(1/2) is easier than what i suggested

  33. tsghernan
    • 2 years ago
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    Yes. I know it. I verified the result with the mathematica soft. I reached it. Thank you a lot

  34. tcarroll010
    • 2 years ago
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    I do believe @cnknd has a point. It looks like \[2\int\limits_{}^{}\frac{ du }{ \sqrt[4]{u} }\]is the integral to be considered. How does that look to you now, @cnknd?

  35. cnknd
    • 2 years ago
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    yea that's what i meant.

  36. tcarroll010
    • 2 years ago
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    and then (8/3)u^(3/4) with the eventual substitution of u in terms of x. Is that what you are getting?

  37. tsghernan
    • 2 years ago
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    You are right ;)

  38. tcarroll010
    • 2 years ago
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    Good catch @cnknd !

  39. tcarroll010
    • 2 years ago
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    So, the answer is derived with a combination of my substitution equation of u = 1 + x^(1/2) (to resolve the other x^(1/2), that part is correct ) and @cnknd 's correction on the exponent on u going from 1/2 to 1/4.

  40. asnaseer
    • 2 years ago
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    nice to see a "team" working towards a solution :)

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