## tsghernan I can not solve this integral one year ago one year ago

1. cnknd

cool story bro

2. tsghernan

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

3. g152xx

@TuringTest is here so have no fear ! :)

4. TuringTest

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

5. cnknd

try rewriting this with fractional powers instead of nested sqrts

6. cnknd

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

7. 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.

8. tcarroll010

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

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

10. TuringTest

@tcarroll010 yes, I agree :)

11. tcarroll010

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

12. cnknd

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

13. tcarroll010

$2\int\limits_{}^{}\frac{ du }{ \sqrt{u} }$

14. tcarroll010

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

15. tcarroll010

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

16. cnknd

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

17. tcarroll010

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

18. cnknd

19. tcarroll010

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

20. cnknd

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

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

22. tcarroll010

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

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

24. tcarroll010

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

25. asnaseer

@cnknd - @tcarroll010 is right - he is making use of this:$u = 1 + x^{1/2}$Then$du = (1/2)x^{-1/2} dx=\frac{dx}{2\sqrt{x}}$

26. cnknd

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

I can solve it. Thank you guys

28. cnknd

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

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

30. asnaseer

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

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

32. cnknd

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

33. tsghernan

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

34. tcarroll010

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

yea that's what i meant.

36. tcarroll010

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

You are right ;)

38. tcarroll010

Good catch @cnknd !

39. tcarroll010

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

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