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Question Number 18233 by Joel577 last updated on 17/Jul/17

$$\mathrm{If}y=\sqrt{x^2+\sqrt{x^2+\sqrt{x^2+\sqrt{...}}}}$$$$\mathrm{find}\int (y+\sqrt{y})dx$$

Commented by alex041103 last updated on 17/Jul/17

$$didyoumeany=\sqrt{x^2+\sqrt{x^2+\sqrt{x^2+...}}}$$

Commented by Joel577 last updated on 17/Jul/17

$$ohyes.thankyou.Ihavecorrectedit$$

Answered by mrW1 last updated on 17/Jul/17

$$y=\sqrt{x^2+\sqrt{x^2+\sqrt{x^2+\sqrt{...}}}}$$$$y^2={\mathrm{x}}^2+\sqrt{x^2+\sqrt{x^2+\sqrt{x^2+\sqrt{...}}}}={\mathrm{x}}^2+\mathrm{y}$$$${\mathrm{y}}^2-\mathrm{y}-{\mathrm{x}}^2=0$$$$\mathrm{y}=\frac{1+\sqrt{1+{4\mathrm{x}}^2}}{2}=\frac{1}{2}+\sqrt{{\left(\frac{1}{2}\right)}^2+{\mathrm{x}}^2}$$$$\mathrm{y}=\mathrm{a}+\sqrt{{\mathrm{a}}^2+{\mathrm{x}}^2}\mathrm{with}\mathrm{a}=\frac{1}{2}$$$$\sqrt{\mathrm{y}}=\sqrt{\mathrm{a}+\sqrt{{\mathrm{a}}^2+{\mathrm{x}}^2}}$$$$\mathrm{y}+\sqrt{\mathrm{y}}=\mathrm{a}+\sqrt{{\mathrm{a}}^2+{\mathrm{x}}^2}+\sqrt{\mathrm{a}+\sqrt{{\mathrm{a}}^2+{\mathrm{x}}^2}}$$$$......$$

Commented by Joel577 last updated on 25/Jul/17

$$thankyouverymuch$$

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