1. **State the problem:** Solve the differential equation $$\sqrt{x+y+1} \frac{dy}{dx} = x + y - 1.$$\n\n2. **Rewrite the equation:** Let $z = x + y$. Then $y = z - x$ and $$\frac{dy}{dx} = \frac{dz}{dx} - 1.$$\nSubstitute into the equation:\n$$\sqrt{z + 1} \left(\frac{dz}{dx} - 1\right) = z - 1.$$\n\n3. **Expand and isolate $\frac{dz}{dx}$:**\n$$\sqrt{z + 1} \frac{dz}{dx} - \sqrt{z + 1} = z - 1,$$\nso\n$$\sqrt{z + 1} \frac{dz}{dx} = z - 1 + \sqrt{z + 1}.$$\n\n4. **Divide both sides by $\sqrt{z + 1}$:**\n$$\frac{dz}{dx} = \frac{z - 1}{\sqrt{z + 1}} + 1.$$\nWrite as\n$$\frac{dz}{dx} = \frac{z - 1}{\sqrt{z + 1}} + \frac{\cancel{\sqrt{z + 1}}}{\cancel{\sqrt{z + 1}}} = \frac{z - 1 + \sqrt{z + 1}}{\sqrt{z + 1}}.$$\n\n5. **Rewrite numerator:** Let’s keep it as is for now and separate variables:\n$$\frac{dz}{dx} = \frac{z - 1 + \sqrt{z + 1}}{\sqrt{z + 1}}.$$\n\n6. **Separate variables:**\n$$\frac{\sqrt{z + 1}}{z - 1 + \sqrt{z + 1}} dz = dx.$$\n\n7. **Substitute $t = \sqrt{z + 1}$, so $z = t^2 - 1$ and $dz = 2t dt$:**\nThe numerator becomes $t$, denominator becomes $(t^2 - 1) - 1 + t = t^2 + t - 2$, so\n$$\frac{t}{t^2 + t - 2} dz = \frac{t}{t^2 + t - 2} 2t dt = \frac{2 t^2}{t^2 + t - 2} dt = dx.$$\n\n8. **Simplify denominator:** Factor denominator:\n$$t^2 + t - 2 = (t + 2)(t - 1).$$\nSo integral becomes\n$$\int \frac{2 t^2}{(t + 2)(t - 1)} dt = \int dx = x + C.$$\n\n9. **Perform partial fraction decomposition:**\nWrite\n$$\frac{2 t^2}{(t + 2)(t - 1)} = \frac{A}{t + 2} + \frac{B}{t - 1}.$$\nMultiply both sides by denominator:\n$$2 t^2 = A(t - 1) + B(t + 2).$$\n\n10. **Solve for A and B:**\nSet $t=1$:\n$$2(1)^2 = A(0) + B(3) \Rightarrow 2 = 3B \Rightarrow B = \frac{2}{3}.$$\nSet $t=-2$:\n$$2(-2)^2 = A(-3) + B(0) \Rightarrow 8 = -3A \Rightarrow A = -\frac{8}{3}.$$\n\n11. **Rewrite integral:**\n$$\int \left(-\frac{8}{3} \frac{1}{t + 2} + \frac{2}{3} \frac{1}{t - 1}\right) dt = x + C.$$\n\n12. **Integrate:**\n$$-\frac{8}{3} \ln|t + 2| + \frac{2}{3} \ln|t - 1| = x + C.$$\n\n13. **Substitute back $t = \sqrt{z + 1} = \sqrt{x + y + 1}$:**\n$$-\frac{8}{3} \ln|\sqrt{x + y + 1} + 2| + \frac{2}{3} \ln|\sqrt{x + y + 1} - 1| = x + C.$$\n\n14. **Final implicit solution:**\n$$-8 \ln|\sqrt{x + y + 1} + 2| + 2 \ln|\sqrt{x + y + 1} - 1| = 3x + C',$$\nwhere $C' = 3C$ is an arbitrary constant.