1. **State the problem:** Simplify the expression $$\frac{x}{x^2-1} - \frac{x-3}{x-1}$$. 2. **Recall the formula and rules:** The denominator $x^2-1$ can be factored using the difference of squares formula: $$x^2-1 = (x-1)(x+1)$$. 3. **Rewrite the expression with factored denominators:** $$\frac{x}{(x-1)(x+1)} - \frac{x-3}{x-1}$$ 4. **Find a common denominator:** The common denominator is $(x-1)(x+1)$. 5. **Rewrite the second fraction to have the common denominator:** $$\frac{x-3}{x-1} = \frac{(x-3)(x+1)}{(x-1)(x+1)}$$ 6. **Combine the fractions:** $$\frac{x}{(x-1)(x+1)} - \frac{(x-3)(x+1)}{(x-1)(x+1)} = \frac{x - (x-3)(x+1)}{(x-1)(x+1)}$$ 7. **Expand the numerator:** $$(x-3)(x+1) = x^2 + x - 3x - 3 = x^2 - 2x - 3$$ 8. **Substitute back and simplify numerator:** $$x - (x^2 - 2x - 3) = x - x^2 + 2x + 3 = -x^2 + 3x + 3$$ 9. **Rewrite the expression:** $$\frac{-x^2 + 3x + 3}{(x-1)(x+1)}$$ 10. **Factor the numerator if possible:** Try factoring $-x^2 + 3x + 3$: $$-x^2 + 3x + 3 = -(x^2 - 3x - 3)$$ The quadratic $x^2 - 3x - 3$ does not factor nicely with integers, so leave as is. 11. **Final simplified expression:** $$\boxed{\frac{-(x^2 - 3x - 3)}{(x-1)(x+1)}}$$ or equivalently $$\boxed{\frac{-x^2 + 3x + 3}{(x-1)(x+1)}}$$ This is the simplified form of the original expression. **Note:** The domain excludes $x=1$ and $x=-1$ because they make the denominator zero.