If$y=\sqrt{(x^2+x+1)+\sqrt{(x^2+x+1)-\sqrt{(x^2+x+1)+\sqrt{(x^2+x+1)\cdots\cdots}}}}$. Then $\displaystyle \int^{3}_{2}ydx$.
Try: writting equation as $$y=\sqrt{(x^2+x+1)+\sqrt{(x^2+x+1)-y}}$$
So $$y^2=x^2+x+1+\sqrt{(x^2+x+1)-y}$$
$$\bigg(y^2-(x^2+x+1)\bigg)^2=(x^2+x+1-y)$$
I did not understand how can i calculate $y$ in terms of $x$
Could some help me to solve it, Thanks
Squaring the radical equation twice is likely to introduce extraneous functional solutions. The doubly-squared equation is difficult to factor "by hand" but Wolfram factors it as
$$ (x-y+1)(x+y)\left(x^2+x-y^2-y+1\right)=0 $$
giving possible solutions
The solution is $y=x+1$ for $x\ge0$ as can be seen by direct substitution into the original equation
$$y=\sqrt{(x^2+x+1)+\sqrt{(x^2+x+1)-y}} $$
so the integral $\int_2^3y\,dy$ is straightforward.
After squaring twice you get four functional solutions, three of which are extraneous. Here is the graph of the last equation stated in the question.
