1. **State the problem:** We need to perform long division on an improper fraction and then express the resulting proper fraction as a sum of partial fractions. 2. **General approach:** - Perform long division if the numerator degree is greater than or equal to the denominator degree. - Write the improper fraction as a polynomial plus a proper fraction. - Decompose the proper fraction into partial fractions. 3. **Example:** Suppose we have $$\frac{2x^2 + 3x + 1}{x^2 - 1}$$. 4. **Perform long division:** Divide $$2x^2 + 3x + 1$$ by $$x^2 - 1$$. $$\frac{2x^2 + 3x + 1}{x^2 - 1} = 2 + \frac{3x + 3}{x^2 - 1}$$ 5. **Factor the denominator of the proper fraction:** $$x^2 - 1 = (x - 1)(x + 1)$$ 6. **Set up partial fractions:** $$\frac{3x + 3}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1}$$ 7. **Multiply both sides by the denominator:** $$3x + 3 = A(x + 1) + B(x - 1)$$ 8. **Expand and group terms:** $$3x + 3 = A x + A + B x - B = (A + B) x + (A - B)$$ 9. **Equate coefficients:** - Coefficient of $$x$$: $$3 = A + B$$ - Constant term: $$3 = A - B$$ 10. **Solve the system:** Add equations: $$3 + 3 = (A + B) + (A - B) \Rightarrow 6 = 2A \Rightarrow A = 3$$ Substitute $$A = 3$$ into $$3 = A + B$$: $$3 = 3 + B \Rightarrow B = 0$$ 11. **Write the partial fraction decomposition:** $$\frac{3x + 3}{(x - 1)(x + 1)} = \frac{3}{x - 1} + \frac{0}{x + 1} = \frac{3}{x - 1}$$ 12. **Final answer:** $$\frac{2x^2 + 3x + 1}{x^2 - 1} = 2 + \frac{3}{x - 1}$$