integral of sqrt(x+sqrt(x+sqrt(x+...))), infinite nested square root

Find the limit of this function as x goes to 0+,

Find the derivative of this function,

integral of infinitely many nested square roots,
fun integral problem,
integral of sqrt(x+sqrt(x+sqrt(x+...)))),
nested radicals,
integration with completing the square,

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blackpenredpen | 曹老師

[Roarshark]

You make solving these fun plus you're so *positive* and *happy* and that's what reaches me ^_^. /THAT'S IT!/

[mlihan]

Are you using a blue pen? I feel cheated lol.

[hamzaaldajani]

I noticed that y^2 is equal to y+x (because it's an infinite serie). You can solve the ecuation y^2 -y-x, taking x as constant by using the cuadratic formula, and you integrate

[AndDiracisHisProphet]

Yo dawg, I see you put the function out of the function so that you didn't have to integrate while you integrate.

[thomasg]

The math is easy to understand, but what i don't get is:
Are there a rabbit and a cat, both named Oreo? 🍪🐇🐈

[TheSenator]

"And we're done" *Continues to write down the integration constant* That's how it's done ;-)

[davidkippy]

All you have to do is implicitly differentiate y^2-y=x, which becomes dy/dx= 1/(2y-1), then you substitute the integrand for y and dx for (2y-1)dy and integrate to get 2/3y^3-1/2y^2

[rosebuster]

Cute rabbit and cute cat! Animals make everything better!

[Fematika]

y(1) is the golden ratio, as you can see on my channel. :)

[ekamjitsingh]

At 1:36 you are squaring both sides so you should open right hand side with modulus

[mathunt]

With the expression y^2-y=x, you have a perfect equation for substitution.

[TheMauror]

Do an integral battle with some Of these crazy functions!

[tanmaywho]

You made me understand so easy. Thanks

[paulwilm]

Thanks, you just helped us with our math contest

[rome]

Some good integral is always satisfying. 😊

[vcvartak]

Expressing y in terms x by adding 1/4 and make it perfect square is nice step

[alhussain]

I noticed that when we substitute 1 in our expression ( 1/2 + sqrt (4x+1) / 2) we get (sqrt(5) + 1)/2 which is the golden ratio. Amazing🔥

[emmeeemm]

Hm. I did this slightly differently. I started with the substitution, got to y=sqrt(x+y), and then to y^2-y=x. But then, I just did u-sub from there, asserting (2y-1)dy = dx. I plugged that into the original integral, and integrated y(2y-1)dy to get (2/3)y^3 - (1/2)y^2 + C, and then I back-substituted the terms of x. So, my answer didn't end up as a closed form.

The method in this video is pretty cool. Most parts of the result are similar to my answer. I bet with a little algebra, it could be shown that they're equivalent.

[pradeepkumarbhatt]

Very interesting problem and good method to solve it.Ramanujan was greatest mathematician giving a lot of excellent theorems and results regarding infinite series.

[murad]

In the 5:30sh, if you set x=0, then y as defined under the radical should also be equal to 0. This implies that you have to consider the -ve sign instead of the +ve (here: y= 1/2 +/-1/2 sqrt(4x+1)