1. **State the problem:** We want to find the partial fraction decomposition of the rational expression $$\frac{9x^2 - x + 14}{(x+1)(x^2+5)}.$$\n\n2. **Formula and setup:** Since the denominator factors into a linear term $(x+1)$ and an irreducible quadratic term $(x^2+5)$, the decomposition has the form:\n$$\frac{9x^2 - x + 14}{(x+1)(x^2+5)} = \frac{A}{x+1} + \frac{Bx + C}{x^2 + 5}$$\nwhere $A$, $B$, and $C$ are constants to be determined.\n\n3. **Multiply both sides by the denominator:**\n$$9x^2 - x + 14 = A(x^2 + 5) + (Bx + C)(x + 1)$$\n\n4. **Expand the right side:**\n$$A x^2 + 5A + Bx^2 + Bx + Cx + C = (A + B) x^2 + (B + C) x + (5A + C)$$\n\n5. **Equate coefficients of like powers of $x$:**\n- Coefficient of $x^2$: $9 = A + B$\n- Coefficient of $x$: $-1 = B + C$\n- Constant term: $14 = 5A + C$\n\n6. **Solve the system:**\nFrom $9 = A + B$, we get $B = 9 - A$.\nSubstitute into $-1 = B + C$: $-1 = (9 - A) + C \Rightarrow C = -1 - 9 + A = A - 10$.\nSubstitute $C$ into $14 = 5A + C$: $14 = 5A + (A - 10) = 6A - 10 \Rightarrow 6A = 24 \Rightarrow A = 4$.\nThen $B = 9 - 4 = 5$ and $C = 4 - 10 = -6$.\n\n7. **Write the final decomposition:**\n$$\frac{9x^2 - x + 14}{(x+1)(x^2+5)} = \frac{4}{x+1} + \frac{5x - 6}{x^2 + 5}.$$