A tricky integral

[elim]

If \(y=\sqrt{x^2+\sqrt{x^2+\sqrt{x^2+\cdots}}}\), find \(\int (y+\sqrt{y})dx\)

[elim]

Sol. Clearly $\small{\displaystyle{y^2 = x^2 + y,\quad y=\frac{\sqrt{4x^2+1}+1}{2},\quad dx = \frac{2y-1}{2\sqrt{y^2-y}}dy \;\overset{y=t^2}{=}\;\frac{2t^2-1}{\sqrt{t^2-1}}dt}}$
$\small{\qquad\begin{align*}I & = \int (y+\sqrt{y})dx =\int \frac{(y +\sqrt{y})(2y -1)}{2\sqrt{y^2 -y}}dy= \int \frac{(t^2+t)(2t^2-1)}{\sqrt{t^2-1}}dt\\& =\frac{1}{12}\bigg((3t+4)(2t^2+1)\sqrt{t^2-1}+3\log\left(t+\sqrt{t^2-1}\right)\bigg)\\& =\frac{1}{12}\bigg((3\sqrt{y}+4)(2y+1)\sqrt{y-1}+3\log\left(\sqrt{y}+\sqrt{y-1}\right)\bigg)\\& =\frac{1}{12}\bigg(\bigg(3x+2\sqrt{2\sqrt{4x^2+1}-2}\bigg)\left(2+\sqrt{4x^2+1}\right)+\frac{3}{2}\log \left(2x+\sqrt{4x^2+1}\right)\bigg)\end{align*}}$

[elim]

Let $y_1 = |x|, \; y_{n+1} = \sqrt{x^2 + y_n}$, then $y_2 \ge y_1,\; y_n \ge y_{n-1}\implies y_{n+1} \ge y_n$. So $0\le y_n \le y_{n+1}\;(\forall n)$
We have $y_1 < |x|+1$. Since $y_n < |x|+1 \implies y_{n+1} = \sqrt{x^2 + y_n} < \sqrt{(|x|+1)^2} = |x|+1$,
so $\{y_n\}$ is bounded. $\displaystyle{y = \lim_{n\to\infty} y_n} = \sqrt{x^2 +\sqrt{x^2+\sqrt{x^2+\cdots}}}$ exists, and $y = \sqrt{x^2 +y}$
Therefore $y = \begin{cases} 0, & x = 0\\ {\small{\dfrac{1+\sqrt{1+4x^2}}{2}}}, & x \ne 0.\end{cases}$ and so $\displaystyle{\int (y + \sqrt{y})dx}$ exists on $\mathbb{R}- \{0\}$.