1. **State the problem:** We are given the differential equation $$\frac{dy}{dx} = \frac{x^2 + 1}{1 - x}$$ and asked to analyze or solve it. 2. **Understand the equation:** This is a first-order ordinary differential equation expressing the derivative of $y$ with respect to $x$. 3. **Rewrite the equation:** We can write it as $$dy = \frac{x^2 + 1}{1 - x} dx$$ to prepare for integration. 4. **Integrate both sides:** To find $y$, integrate the right side with respect to $x$: $$y = \int \frac{x^2 + 1}{1 - x} dx + C$$ 5. **Simplify the integrand:** Perform polynomial division or rewrite the fraction: Divide $x^2 + 1$ by $1 - x$: $$\frac{x^2 + 1}{1 - x} = -x - 1 - \frac{x}{1 - x}$$ Explanation: Because $$x^2 + 1 = (1 - x)(-x - 1) + x$$ 6. **Rewrite the integral:** $$y = \int (-x - 1) dx - \int \frac{x}{1 - x} dx + C$$ 7. **Integrate the first part:** $$\int (-x - 1) dx = -\frac{x^2}{2} - x$$ 8. **Integrate the second part:** Rewrite $$\frac{x}{1 - x} = \frac{1 - (1 - x)}{1 - x} = \frac{1}{1 - x} - 1$$ So, $$- \int \frac{x}{1 - x} dx = - \int \left( \frac{1}{1 - x} - 1 \right) dx = - \int \frac{1}{1 - x} dx + \int 1 dx$$ 9. **Integrate each term:** $$- \int \frac{1}{1 - x} dx = -(-\ln|1 - x|) = \ln|1 - x|$$ and $$\int 1 dx = x$$ 10. **Combine all parts:** $$y = -\frac{x^2}{2} - x + \ln|1 - x| + x + C = -\frac{x^2}{2} + \ln|1 - x| + C$$ 11. **Final answer:** $$\boxed{y = -\frac{x^2}{2} + \ln|1 - x| + C}$$ This is the general solution to the differential equation.