1. **State the problem:** Solve the equation $$\frac{1}{x^2 - x} + \frac{1}{x} = 1.$$\n\n2. **Rewrite the equation and identify the domain:** Note that $$x^2 - x = x(x-1),$$ so the denominators are zero at $$x=0$$ and $$x=1,$$ which are excluded from the domain.\n\n3. **Find a common denominator:** The common denominator is $$x(x-1).$$ Rewrite each term with this denominator:\n$$\frac{1}{x^2 - x} = \frac{1}{x(x-1)}, \quad \frac{1}{x} = \frac{x-1}{x(x-1)}, \quad 1 = \frac{x(x-1)}{x(x-1)}.$$\n\n4. **Rewrite the equation with common denominator:**\n$$\frac{1}{x(x-1)} + \frac{x-1}{x(x-1)} = \frac{x(x-1)}{x(x-1)}.$$\n\n5. **Combine the left side:**\n$$\frac{1 + (x-1)}{x(x-1)} = \frac{x(x-1)}{x(x-1)}.$$\nSimplify numerator:\n$$1 + (x-1) = x.$$\nSo the equation becomes:\n$$\frac{x}{x(x-1)} = \frac{x(x-1)}{x(x-1)}.$$\n\n6. **Simplify the left side:**\n$$\frac{x}{x(x-1)} = \frac{\cancel{x}}{\cancel{x}(x-1)} = \frac{1}{x-1}.$$\nSo the equation is now:\n$$\frac{1}{x-1} = \frac{x(x-1)}{x(x-1)} = 1.$$\n\n7. **Solve the simplified equation:**\n$$\frac{1}{x-1} = 1 \implies 1 = x - 1 \implies x = 2.$$\n\n8. **Check domain restrictions:**\n$$x=2$$ is not excluded (not 0 or 1), so it is a valid solution.\n\n**Final answer:** $$\boxed{2}.$$