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Question Number 80144 by behi83417@gmail.com last updated on 31/Jan/20
solve for x:  (((√x)+1)/( (√(x+1))))+ax^2 =x(a^2 +1)      [a∈R]
$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{x}}: \\ $$$$\frac{\sqrt{\boldsymbol{\mathrm{x}}}+\mathrm{1}}{\:\sqrt{\boldsymbol{\mathrm{x}}+\mathrm{1}}}+\boldsymbol{\mathrm{ax}}^{\mathrm{2}} =\boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\mathrm{1}\right)\:\:\:\:\:\:\left[\boldsymbol{\mathrm{a}}\in\boldsymbol{\mathrm{R}}\right] \\ $$
Commented by john santu last updated on 31/Jan/20
let (√(x+1)) = sec t  x+1 = sec^2 t⇒x = tan^2 t  ((tan t+1)/(sec t))+a(tan^4 t)=(a^2 +1)tan^2 t  cos t(tan t+1)+atan^4 t=(a^2 +1)tan^2 t
$${let}\:\sqrt{{x}+\mathrm{1}}\:=\:\mathrm{sec}\:{t} \\ $$$${x}+\mathrm{1}\:=\:\mathrm{sec}\:^{\mathrm{2}} {t}\Rightarrow{x}\:=\:\mathrm{tan}\:^{\mathrm{2}} {t} \\ $$$$\frac{\mathrm{tan}\:{t}+\mathrm{1}}{\mathrm{sec}\:{t}}+{a}\left(\mathrm{tan}\:^{\mathrm{4}} {t}\right)=\left({a}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{tan}\:^{\mathrm{2}} {t} \\ $$$$\mathrm{cos}\:{t}\left(\mathrm{tan}\:{t}+\mathrm{1}\right)+{a}\mathrm{tan}\:^{\mathrm{4}} {t}=\left({a}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{tan}\:^{\mathrm{2}} {t} \\ $$

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