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tsghernan
I can not solve this integral
\[\int\limits_{1}^{9} \frac{ 1 }{ \sqrt{x*\sqrt{1+\sqrt{x}}} } \delta x \] (The integral is between 9 and 1
@TuringTest is here so have no fear ! :)
Why do some people think I am a math demi-god? I have no idea.... at least not yet.
try rewriting this with fractional powers instead of nested sqrts
then do a u-substitution where u = sqrt(x) (i'm just eyeballing this but it seems like this should work)
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.
No, substitute u = 1 + x^(1/2) Then du = (1/2)x^(-1/2) dx and then easy to solve from that point.
heh, same thing, just off by 1 (du is still the same)
@tcarroll010 yes, I agree :)
@cnknd , it is necessary to isolate that term of "1" into u along with the x^(1/2)
i don't think it's all that difficult to integrate (1+u)^(-1/4) rather than u^(-1/4)
\[2\int\limits_{}^{}\frac{ du }{ \sqrt{u} }\]
\[= 4\sqrt{u}\]Now, just put the expression for u back in.
\[4\sqrt{1 + \sqrt{x}}\]And you're done.
@tcarroll010 you forgot the a 2nd sqrt. it's \[2\int\limits_{2}^{4}\frac{ du }{ \sqrt{\sqrt{u}} }\]
Did not forget. Go back and look at the du equation. sqrt(x) is already in it.
They get resolved in the du equation. Simply take the derivative of my answer and you'll see.
the first sqrt from the original expression covers both the x and the sqrt(1+sqrt(x)) unless my eyes are deceiving me.
@cnknd - @tcarroll010 is right, take a look again.
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.
original integrand was: \[\frac{ 1 }{ \sqrt{x \sqrt{1+\sqrt{x}}} }\]
yes, which is the same as my re-write.
@cnknd - @tcarroll010 is right - he is making use of this:\[u = 1 + x^{1/2}\]Then\[du = (1/2)x^{-1/2} dx=\frac{dx}{2\sqrt{x}}\]
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
I can solve it. Thank you guys
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
tcarroll010 was right. The best way is substitute u = 1 + x^(1/2)
oh - sorry @cnknd - I thought you were wondering where the \(\sqrt{x}\) went. you are right - there should be an extra sqrt there.
@tsghernan - please take note of what @cnknd was saying as well - there is an extra sqrt to put in
thank you sir moderator :) and yea u = 1 + x^(1/2) is easier than what i suggested
Yes. I know it. I verified the result with the mathematica soft. I reached it. Thank you a lot
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?
and then (8/3)u^(3/4) with the eventual substitution of u in terms of x. Is that what you are getting?
Good catch @cnknd !
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.
nice to see a "team" working towards a solution :)