Express (x³-x²-x-4)/(x(x²+2)) in partial fractions.

The given expression is $\frac{x^3 - x^2 - x - 4}{x(x^2 + 2)} = \frac{x^3 - x^2 - x - 4}{x^3 + 2x}$.

$$\frac{x^3 - x^2 - x - 4}{x^3 + 2x} = 1 + \frac{(x^3 - x^2 - x - 4) - (x^3 + 2x)}{x^3 + 2x}$$ $$= 1 + \frac{-x^2 - 3x - 4}{x^3 + 2x}$$

Let the proper fraction be $\frac{-x^2 - 3x - 4}{x(x^2 + 2)}$. Since $x$ is a linear factor and $x^2 + 2$ is an irreducible quadratic factor, the decomposition is:

$$\frac{-x^2 - 3x - 4}{x(x^2 + 2)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 2}$$

Multiply both sides by $x(x^2 + 2)$:

$$-x^2 - 3x - 4 = A(x^2 + 2) + (Bx + C)x$$

Expand the right side:

$$-x^2 - 3x - 4 = Ax^2 + 2A + Bx^2 + Cx$$

Group terms by powers of $x$:

$$-x^2 - 3x - 4 = (A + B)x^2 + Cx + 2A$$

Equate the coefficients of corresponding powers of $x$:
For $x^2$: $A + B = -1$ (Equation 1)
For $x$: $C = -3$ (Equation 2)
For the constant term: $2A = -4$ (Equation 3)

From Equation 3:

$$2A = -4 \implies A = -2$$

Substitute $A = -2$ into Equation 1:

$$-2 + B = -1 \implies B = -1 + 2 \implies B = 1$$

So, we have $A = -2$, $B = 1$, and $C = -3$.

$$\frac{x^3 - x^2 - x - 4}{x(x^2 + 2)} = 1 + \frac{-2}{x} + \frac{x - 3}{x^2 + 2}$$

The final answer is $1 - \frac{2}{x} + \frac{x - 3}{x^2 + 2}$.