Compute the indefinite integral
$$\displaystyle \int\sqrt{x\sqrt{x\sqrt{x\cdots\sqrt{x}}}}~~dx=\displaystyle \int\sqrt{x\underset{n~\text{times}}{\underbrace{\sqrt{x\sqrt{x\sqrt{x...\sqrt{x}}}}}}}~~dx$$ where $x$ is repeated $n$ times.
My Attempt (I hope I have succeeded in the solution)
$1)$ We put $$\displaystyle y_{n}=\sqrt{x\sqrt{x\sqrt{x...\sqrt{x}}}}$$
so we have
$$\displaystyle y_{n}=x^{\frac{2^{n}-1}{2^{n}}}=x^{1-\frac{1}{2^{n}}}$$
$$\displaystyle \int y_{n}dx=\frac{x^{2-\frac{1}{2^{n}}}}{2-\frac{1}{2^{n}}}+c$$
$2)$ Important note $$\displaystyle \lim_{n \to \infty }\displaystyle\displaystyle\int\sqrt{x\underset{n}{\underbrace{\sqrt{x\sqrt{x\sqrt{x...\sqrt{x}}}}}}}~dx=\frac{x^{2}}{2}+c $$
You have all my respect and appreciation. Thank you!
Note that:$$\sqrt{x\sqrt{x\sqrt{x...\sqrt{x}}}}=x^{\frac{1}{2}} x^{\frac{1}4} x^{\frac{1}8}\cdots x^{\frac{1}{2^n}}=x^{\frac{1}2+\frac{1}4+\cdots+\frac{1}{2^n}}=x^{1-\frac{1}{2^n}}$$
$$\int \sqrt{x\sqrt{x\sqrt{x...\sqrt{x}}}}~~ dx=\frac{1}{2-\frac{1}{2^n}}x^{2-\frac{1}{2^n}}+C$$