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