I can not solve this integral

- tsghernan

I can not solve this integral

- Stacey Warren - Expert brainly.com

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- katieb

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- anonymous

cool story bro

- tsghernan

\[\int\limits_{1}^{9} \frac{ 1 }{ \sqrt{x*\sqrt{1+\sqrt{x}}} } \delta x \] (The integral is between 9 and 1

- g152xx

@TuringTest is here so have no fear ! :)

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## More answers

- TuringTest

Why do some people think I am a math demi-god? I have no idea.... at least not yet.

- anonymous

try rewriting this with fractional powers instead of nested sqrts

- anonymous

then do a u-substitution where u = sqrt(x)
(i'm just eyeballing this but it seems like this should work)

- TuringTest

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.

- anonymous

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

- anonymous

heh, same thing, just off by 1 (du is still the same)

- TuringTest

@tcarroll010 yes, I agree :)

- anonymous

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

- anonymous

i don't think it's all that difficult to integrate (1+u)^(-1/4) rather than u^(-1/4)

- anonymous

\[2\int\limits_{}^{}\frac{ du }{ \sqrt{u} }\]

- anonymous

\[= 4\sqrt{u}\]Now, just put the expression for u back in.

- anonymous

\[4\sqrt{1 + \sqrt{x}}\]And you're done.

- anonymous

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

- anonymous

Did not forget. Go back and look at the du equation. sqrt(x) is already in it.

- anonymous

original eqn had 3 sqrts

- anonymous

They get resolved in the du equation. Simply take the derivative of my answer and you'll see.

- anonymous

the first sqrt from the original expression covers both the x and the sqrt(1+sqrt(x))
unless my eyes are deceiving me.

- g152xx

@cnknd - @tcarroll010 is right, take a look again.

- anonymous

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.

- anonymous

original integrand was:
\[\frac{ 1 }{ \sqrt{x \sqrt{1+\sqrt{x}}} }\]

- anonymous

yes, which is the same as my re-write.

- asnaseer

@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

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

- tsghernan

I can solve it. Thank you guys

- anonymous

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

- tsghernan

tcarroll010 was right. The best way is substitute u = 1 + x^(1/2)

- asnaseer

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

- asnaseer

@tsghernan - please take note of what @cnknd was saying as well - there is an extra sqrt to put in

- anonymous

thank you sir moderator :)
and yea u = 1 + x^(1/2) is easier than what i suggested

- tsghernan

Yes. I know it. I verified the result with the mathematica soft. I reached it. Thank you a lot

- anonymous

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?

- anonymous

yea that's what i meant.

- anonymous

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

- tsghernan

You are right ;)

- anonymous

Good catch @cnknd !

- anonymous

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.

- asnaseer

nice to see a "team" working towards a solution :)

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