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tsghernan
 4 years ago
I can not solve this integral
tsghernan
 4 years ago
I can not solve this integral

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tsghernan
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{1}^{9} \frac{ 1 }{ \sqrt{x*\sqrt{1+\sqrt{x}}} } \delta x \] (The integral is between 9 and 1

g152xx
 4 years ago
Best ResponseYou've already chosen the best response.0@TuringTest is here so have no fear ! :)

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0Why do some people think I am a math demigod? I have no idea.... at least not yet.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0try rewriting this with fractional powers instead of nested sqrts

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then do a usubstitution where u = sqrt(x) (i'm just eyeballing this but it seems like this should work)

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0you 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No, substitute u = 1 + x^(1/2) Then du = (1/2)x^(1/2) dx and then easy to solve from that point.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0heh, same thing, just off by 1 (du is still the same)

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0@tcarroll010 yes, I agree :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@cnknd , it is necessary to isolate that term of "1" into u along with the x^(1/2)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i don't think it's all that difficult to integrate (1+u)^(1/4) rather than u^(1/4)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[2\int\limits_{}^{}\frac{ du }{ \sqrt{u} }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[= 4\sqrt{u}\]Now, just put the expression for u back in.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[4\sqrt{1 + \sqrt{x}}\]And you're done.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@tcarroll010 you forgot the a 2nd sqrt. it's \[2\int\limits_{2}^{4}\frac{ du }{ \sqrt{\sqrt{u}} }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Did not forget. Go back and look at the du equation. sqrt(x) is already in it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0original eqn had 3 sqrts

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0They get resolved in the du equation. Simply take the derivative of my answer and you'll see.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the first sqrt from the original expression covers both the x and the sqrt(1+sqrt(x)) unless my eyes are deceiving me.

g152xx
 4 years ago
Best ResponseYou've already chosen the best response.0@cnknd  @tcarroll010 is right, take a look again.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Simply 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0original integrand was: \[\frac{ 1 }{ \sqrt{x \sqrt{1+\sqrt{x}}} }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes, which is the same as my rewrite.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0@cnknd  @tcarroll010 is right  he is making use of this:\[u = 1 + x^{1/2}\]Then\[du = (1/2)x^{1/2} dx=\frac{dx}{2\sqrt{x}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let \[A = \sqrt{1+\sqrt{x}}\] the original integrand would be: \[\frac{ 1 }{ \sqrt{xA} }\] and your rewrite is: \[\frac{ 1 }{ \sqrt{x}*A }\] im pretty sure those are different

tsghernan
 4 years ago
Best ResponseYou've already chosen the best response.0I can solve it. Thank you guys

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if 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

tsghernan
 4 years ago
Best ResponseYou've already chosen the best response.0tcarroll010 was right. The best way is substitute u = 1 + x^(1/2)

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0oh  sorry @cnknd  I thought you were wondering where the \(\sqrt{x}\) went. you are right  there should be an extra sqrt there.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0@tsghernan  please take note of what @cnknd was saying as well  there is an extra sqrt to put in

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thank you sir moderator :) and yea u = 1 + x^(1/2) is easier than what i suggested

tsghernan
 4 years ago
Best ResponseYou've already chosen the best response.0Yes. I know it. I verified the result with the mathematica soft. I reached it. Thank you a lot

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea that's what i meant.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and then (8/3)u^(3/4) with the eventual substitution of u in terms of x. Is that what you are getting?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So, 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.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0nice to see a "team" working towards a solution :)
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