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Question-201555




Question Number 201555 by Simurdiera last updated on 08/Dec/23
Answered by mr W last updated on 09/Dec/23
let u=x+y  (du/dx)=1+(dy/dx) ⇒(dy/dx)=(du/dx)−1  ⇒(√(u+1))((du/dx)−1)=(√(u−1))  ⇒(du/dx)=((√(u−1))/( (√(u+1))))+1  ⇒((√(u+1))/( (√(u+1))+(√(u−1))))du=dx  ⇒((((√(u+1))−(√(u−1)))(√(u+1)))/( 2))du=dx  ⇒(((u+1−(√(u^2 −1)))du)/( 2))=dx  ⇒∫(u+1−(√(u^2 −1)))du=2∫dx  ⇒(u^2 /2)+u−((u(√(u^2 −1))−ln ((√(u^2 −1))+u))/2)=2x+C  ⇒u^2 +2u+1−u(√(u^2 −1))+ln ((√(u^2 −1))+u)−4x=C  ⇒(u+1)^2 −u(√(u^2 −1))+ln ((√(u^2 −1))+u)−4x=C  ⇒(x+y+1)^2 −(x+y)(√((x+y)^2 −1))+ln ((√((x+y)^2 −1))+x+y)−4x=C
$${let}\:{u}={x}+{y} \\ $$$$\frac{{du}}{{dx}}=\mathrm{1}+\frac{{dy}}{{dx}}\:\Rightarrow\frac{{dy}}{{dx}}=\frac{{du}}{{dx}}−\mathrm{1} \\ $$$$\Rightarrow\sqrt{{u}+\mathrm{1}}\left(\frac{{du}}{{dx}}−\mathrm{1}\right)=\sqrt{{u}−\mathrm{1}} \\ $$$$\Rightarrow\frac{{du}}{{dx}}=\frac{\sqrt{{u}−\mathrm{1}}}{\:\sqrt{{u}+\mathrm{1}}}+\mathrm{1} \\ $$$$\Rightarrow\frac{\sqrt{{u}+\mathrm{1}}}{\:\sqrt{{u}+\mathrm{1}}+\sqrt{{u}−\mathrm{1}}}{du}={dx} \\ $$$$\Rightarrow\frac{\left(\sqrt{{u}+\mathrm{1}}−\sqrt{{u}−\mathrm{1}}\right)\sqrt{{u}+\mathrm{1}}}{\:\mathrm{2}}{du}={dx} \\ $$$$\Rightarrow\frac{\left({u}+\mathrm{1}−\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}\right){du}}{\:\mathrm{2}}={dx} \\ $$$$\Rightarrow\int\left({u}+\mathrm{1}−\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}\right){du}=\mathrm{2}\int{dx} \\ $$$$\Rightarrow\frac{{u}^{\mathrm{2}} }{\mathrm{2}}+{u}−\frac{{u}\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}−\mathrm{ln}\:\left(\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}+{u}\right)}{\mathrm{2}}=\mathrm{2}{x}+{C} \\ $$$$\Rightarrow{u}^{\mathrm{2}} +\mathrm{2}{u}+\mathrm{1}−{u}\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}+\mathrm{ln}\:\left(\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}+{u}\right)−\mathrm{4}{x}={C} \\ $$$$\Rightarrow\left({u}+\mathrm{1}\right)^{\mathrm{2}} −{u}\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}+\mathrm{ln}\:\left(\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}+{u}\right)−\mathrm{4}{x}={C} \\ $$$$\Rightarrow\left({x}+{y}+\mathrm{1}\right)^{\mathrm{2}} −\left({x}+{y}\right)\sqrt{\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{1}}+\mathrm{ln}\:\left(\sqrt{\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{1}}+{x}+{y}\right)−\mathrm{4}{x}={C} \\ $$

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