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Question Number 201555 by Simurdiera last updated on 08/Dec/23

Answered by mr W last updated on 09/Dec/23

$$letu=x+y$$$$\frac{du}{dx}=1+\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{du}{dx}-1$$$$\Rightarrow \sqrt{u+1}(\frac{du}{dx}-1)=\sqrt{u-1}$$$$\Rightarrow \frac{du}{dx}=\frac{\sqrt{u-1}}{\sqrt{u+1}}+1$$$$\Rightarrow \frac{\sqrt{u+1}}{\sqrt{u+1}+\sqrt{u-1}}du=dx$$$$\Rightarrow \frac{(\sqrt{u+1}-\sqrt{u-1})\sqrt{u+1}}{2}du=dx$$$$\Rightarrow \frac{(u+1-\sqrt{u^2-1})du}{2}=dx$$$$\Rightarrow \int (u+1-\sqrt{u^2-1})du=2\int dx$$$$\Rightarrow \frac{u^2}{2}+u-\frac{u\sqrt{u^2-1}-\ln (\sqrt{u^2-1}+u)}{2}=2x+C$$$$\Rightarrow u^2+2u+1-u\sqrt{u^2-1}+\ln (\sqrt{u^2-1}+u)-4x=C$$$$\Rightarrow {(u+1)}^2-u\sqrt{u^2-1}+\ln (\sqrt{u^2-1}+u)-4x=C$$$$\Rightarrow {(x+y+1)}^2-(x+y)\sqrt{{(x+y)}^2-1}+\ln (\sqrt{{(x+y)}^2-1}+x+y)-4x=C$$

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