Question

I can not solve this integral

Answer 1

cool story bro

Answer 2

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

Answer 3

@TuringTest is here so have no fear ! :)

Answer 4

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

Answer 5

try rewriting this with fractional powers instead of nested sqrts

Answer 6

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

Answer 7

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.

Answer 8

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

Answer 9

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

Answer 10

@tcarroll010 yes, I agree :)

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

original eqn had 3 sqrts

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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.

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

I can solve it. Thank you guys

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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?

Answer 35

yea that's what i meant.

Answer 36

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

Answer 37

You are right ;)

Answer 38

Good catch @cnknd !

Answer 39

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.

Answer 40

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