Question Number 54797 by afachri last updated on 11/Feb/19
$$ \\ $$$$ \\ $$$$\:\:\:\int\:\:\sqrt{\:\boldsymbol{{x}}\:+\:\sqrt{\:\boldsymbol{{x}}\:+\:\sqrt{\boldsymbol{{x}}\:+\:\sqrt{\:…..}}}}\:\:\boldsymbol{{dx}}\:\:=\:\:\:? \\ $$$$ \\ $$
Answered by Smail last updated on 11/Feb/19
$${y}=\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}…}}}}} \\ $$$${y}^{\mathrm{2}} ={x}+{y}\Leftrightarrow\left({y}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ={x}+\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${y}=\frac{\mathrm{1}+\sqrt{\mathrm{4}{x}+\mathrm{1}}}{\mathrm{2}} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\sqrt{\mathrm{4}{x}+\mathrm{1}}+\mathrm{1}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{6}}\sqrt{\left(\mathrm{4}{x}+\mathrm{1}\right)^{\mathrm{3}} }+{x}\right)+{C} \\ $$$$ \\ $$
Commented by afachri last updated on 11/Feb/19
$$\boldsymbol{\mathrm{well}}\:\boldsymbol{\mathrm{done}}\:\boldsymbol{\mathrm{Sir}}.\:\boldsymbol{\mathrm{thank}}\:\boldsymbol{\mathrm{you}} \\ $$
Commented by Smail last updated on 11/Feb/19
$${You}\:{are}\:{welcome} \\ $$