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$$\begin{gathered} \mathrm{Given} \\ \sqrt{1+\mathrm{x}+\mathrm{y}}\frac{\mathrm{dy}}{\mathrm{dx}}=\left(\mathrm{x}+\mathrm{y}-1\right) \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\left(\mathrm{x}+\mathrm{y}-1\right)}{\sqrt{1+\mathrm{x}+\mathrm{y}}} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\left(\mathrm{x}+\mathrm{y}+1-2\right)}{\sqrt{1+\mathrm{x}+\mathrm{y}}}=\frac{\mathrm{x}+\mathrm{y}+1}{\sqrt{\mathrm{x}+\mathrm{y}+1}}-\frac{2}{\sqrt{\mathrm{x}+\mathrm{y}+1}} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{\mathrm{x}+\mathrm{y}+1}-\frac{2}{\sqrt{\mathrm{x}+\mathrm{y}+1}} ....\left(1\right) \\ \mathrm{let} \sqrt{\mathrm{x}+\mathrm{y}+1}=\mathrm{u} \\ \mathrm{x}+\mathrm{y}+1={\mathrm{u}}^2 \\ \mathrm{differentiating} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \\ 1+\frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{u}\frac{\mathrm{du}}{\mathrm{dx}} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{u}\frac{\mathrm{du}}{\mathrm{dx}}-1 \\ \mathrm{Now}, \mathrm{from} \left(1\right), \mathrm{we} \mathrm{get} \\ 2\mathrm{u}\frac{\mathrm{du}}{\mathrm{dx}}-1=\mathrm{u}-\frac{2}{\mathrm{u}} \\ \Rightarrow 2\mathrm{u}\frac{\mathrm{du}}{\mathrm{dx}}=\frac{{\mathrm{u}}^2-2}{\mathrm{u}}+1 \\ \Rightarrow 2\mathrm{u}\frac{\mathrm{du}}{\mathrm{dx}}=\frac{{\mathrm{u}}^2-2+\mathrm{u}}{\mathrm{u}} \\ \Rightarrow \left(\frac{2{\mathrm{u}}^2}{{\mathrm{u}}^2-2+\mathrm{u}}\right)\mathrm{du}=\mathrm{dx} \\ \Rightarrow \left(\frac{2{\mathrm{u}}^2}{{\mathrm{u}}^2+2\mathrm{u}-\mathrm{u}-2}\right)\mathrm{du}=\mathrm{dx} \\ \Rightarrow \left(\frac{2{\mathrm{u}}^2}{\mathrm{u}\left(\mathrm{u}+2\right)-1\left(\mathrm{u}+2\right)}\right)\mathrm{du}=\mathrm{dx} \\ \Rightarrow \left(\frac{2{\mathrm{u}}^2}{\left(\mathrm{u}+2\right)\left(\mathrm{u}-1\right)}\right)\mathrm{du}=\mathrm{dx} \\ \Rightarrow \left(\frac{2\mathrm{u}}{3\left(\mathrm{u}-1\right)}+\frac{4\mathrm{u}}{3\left(\mathrm{u}+2\right)}\right)\mathrm{du}=\mathrm{dx} \\ \Rightarrow \frac{2}{3}\left(\frac{\mathrm{u}}{\mathrm{u}-1}\right)\mathrm{du}+\frac{4}{3}\left(\frac{\mathrm{u}}{\mathrm{u}+2}\right)\mathrm{du}=\mathrm{dx} \\ \Rightarrow \frac{2}{3}\left(\frac{\mathrm{u}-1+1}{\mathrm{u}-1}\right)\mathrm{du}+\frac{4}{3}\left(\frac{\mathrm{u}+2-2}{\mathrm{u}+2}\right)\mathrm{du}=\mathrm{dx} \\ \Rightarrow \frac{2}{3}\left(1+\frac{1}{\mathrm{u}-1}\right)\mathrm{du}+\frac{4}{3}\left(1-\frac{2}{\mathrm{u}+2}\right)\mathrm{du}=\mathrm{dx} \\ \mathrm{integrating} \mathrm{both} \mathrm{sides}: \\ \Rightarrow \int \frac{2}{3}\left(1+\frac{1}{\mathrm{u}-1}\right)\mathrm{du}+\int \frac{4}{3}\left(1-\frac{2}{\mathrm{u}+2}\right)\mathrm{du}=\int \mathrm{dx} \\ \Rightarrow \frac{2}{3}\left(\mathrm{u}+\ln \left|\mathrm{u}-1\right|\right)+\frac{4}{3}\left(\mathrm{u}-2\ln \left|\mathrm{u}+2\right|\right)=\mathrm{x}+{\mathrm{C}}_1 \\ \Rightarrow 6\mathrm{u}+2\ln \left|\mathrm{u}-1\right|-8\ln \left|\mathrm{u}+2\right|=3\mathrm{x}+\mathrm{C} \\ \mathrm{put} \mathrm{u}=\sqrt{\mathrm{x}+\mathrm{y}+1} \\ \Rightarrow \boxed{6\sqrt{\mathrm{x}+\mathrm{y}+1}+2\ln \left|\sqrt{\mathrm{x}+\mathrm{y}+1}-1\right|-8\ln \left|\sqrt{\mathrm{x}+\mathrm{y}+1}+2\right|=3\mathrm{x}+\mathrm{C}} \end{gathered}$$

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