1. **State the problem:** We are given the equation $$\frac{4x^3 + 12x^2 - x - 4}{4x^2 - 1} = Ax + B + \frac{C}{2x - 1} + \frac{D}{2x + 1}$$ where $A$, $B$, $C$, and $D$ are constants to be found. 2. **Understand the denominator factorization:** Note that $$4x^2 - 1 = (2x - 1)(2x + 1)$$ which matches the denominators in the partial fractions. 3. **Multiply both sides by the denominator to clear fractions:** $$4x^3 + 12x^2 - x - 4 = (Ax + B)(4x^2 - 1) + C(2x + 1) + D(2x - 1)$$ 4. **Expand the right side:** First expand $(Ax + B)(4x^2 - 1)$: $$= Ax \cdot 4x^2 + B \cdot 4x^2 - Ax - B = 4A x^3 + 4B x^2 - A x - B$$ Now add the other terms: $$4A x^3 + 4B x^2 - A x - B + 2C x + C + 2D x - D$$ Group like terms: $$4A x^3 + 4B x^2 + (-A + 2C + 2D) x + (-B + C - D)$$ 5. **Equate coefficients of corresponding powers of $x$ on both sides:** From the left side: $4x^3 + 12x^2 - x - 4$ - Coefficient of $x^3$: $4 = 4A$ - Coefficient of $x^2$: $12 = 4B$ - Coefficient of $x$: $-1 = -A + 2C + 2D$ - Constant term: $-4 = -B + C - D$ 6. **Solve for $A$ and $B$ first:** $$4 = 4A \implies A = 1$$ $$12 = 4B \implies B = 3$$ 7. **Substitute $A$ and $B$ into the remaining equations:** $$-1 = -1 + 2C + 2D \implies 0 = 2C + 2D \implies C + D = 0$$ $$-4 = -3 + C - D \implies -4 + 3 = C - D \implies -1 = C - D$$ 8. **Solve the system:** From $C + D = 0$, we get $D = -C$. Substitute into $C - D = -1$: $$C - (-C) = -1 \implies 2C = -1 \implies C = -\frac{1}{2}$$ Then $$D = -C = \frac{1}{2}$$ **Final answer:** $$A = 1, \quad B = 3, \quad C = -\frac{1}{2}, \quad D = \frac{1}{2}$$