[tex] \Large \bold{PROBLEM:} [/tex]
[tex] \boxed{\begin{array}{l} \text{Evaluate:} \\ \\ \quad \displaystyle \int_0^{2021} \dfrac{\sqrt{x+\sqrt{x+\sqrt{x+... }}}}{1+\sqrt{x-\sqrt{x-\sqrt{x-...}}}} dx \quad \\ \: \end{array}} [/tex]
[tex] \Large \bold{SOLUTION:} [/tex]
[tex] \begin{array}{l} \textsf{To simplify the }\underline{\textsf{numerator}} \\ \\ \textsf{Let }u = \sqrt{x+\sqrt{x+\sqrt{x+... }}} \\ \\ \textsf{Square both sides.} \\ \\ u^2 = x + \sqrt{x+\sqrt{x+\sqrt{x+... }}} \\ \\ u^2 = x + u \\ \\ u^2 - u = x \\ \\ \textsf{By completing the square,} \\ \\ u^2 - u + \dfrac{1}{4} = x + \dfrac{1}{4} \\ \\ \left(u - \dfrac{1}{2}\right)^2 = x + \dfrac{1}{4} \\ \\ u - \dfrac{1}{2} = \pm \sqrt{x + \dfrac{1}{4}} \\ \\ u = \dfrac{1}{2} \pm \sqrt{x + \dfrac{1}{4}} \\ \: \end{array} [/tex]
[tex] \begin{array}{l} \textsf{Simplifying the }\underline{\textsf{denominator}} \\ \\ \textsf{Let }v = \sqrt{x-\sqrt{x-\sqrt{x-...}}} \\ \\ \textsf{Square both sides.} \\ \\ v^2 = x - \sqrt{x-\sqrt{x-\sqrt{x-...}}} \\ \\ v^2 = x - v \\ \\ v^2 + v = x \\ \\ \textsf{By completing the square,} \\ \\ v^2 + v + \dfrac{1}{4} = x + \dfrac{1}{4} \\ \\ \left(v + \dfrac{1}{2}\right)^2 = x + \dfrac{1}{4} \\ \\ v + \dfrac{1}{2} = \pm \sqrt{x + \dfrac{1}{4}} \\ \\ v = -\dfrac{1}{2} \pm \sqrt{x + \dfrac{1}{4}} \end{array} [/tex]
[tex] \begin{array}{l} \textsf{The integral becomes} \\ \\ \implies \displaystyle \int_0^{2021} \dfrac{\frac{1}{2} \pm \sqrt{x + \frac{1}{4}}}{1 -\frac{1}{2} \pm \sqrt{x + \frac{1}{4}}}dx \\ \\ \implies \displaystyle \int_0^{2021} \dfrac{\frac{1}{2} \pm \sqrt{x + \frac{1}{4}}}{\frac{1}{2} \pm \sqrt{x + \frac{1}{4}}}dx \\ \\ \implies \displaystyle \int_0^{2021} dx = x \Big |_0^{2021} = \boxed{2021} \end{array} [/tex]
