1. The problem is to find the partial fraction decomposition of the rational expression $$\frac{x^2 + 4}{(x+1)(x^2 + 2)}.$$\n\n2. The formula for partial fraction decomposition when the denominator has a linear factor and an irreducible quadratic factor is:\n$$\frac{P(x)}{(x+1)(x^2+2)} = \frac{A}{x+1} + \frac{Bx + C}{x^2 + 2}$$\nwhere $A$, $B$, and $C$ are constants to be determined.\n\n3. Multiply both sides by the denominator $(x+1)(x^2+2)$ to clear the fractions:\n$$x^2 + 4 = A(x^2 + 2) + (Bx + C)(x + 1)$$\n\n4. Expand the right side:\n$$x^2 + 4 = A x^2 + 2A + B x^2 + B x + C x + C$$\n$$x^2 + 4 = (A + B) x^2 + (B + C) x + (2A + C)$$\n\n5. Equate coefficients of like powers of $x$ on both sides:\n- Coefficient of $x^2$: $1 = A + B$\n- Coefficient of $x$: $0 = B + C$\n- Constant term: $4 = 2A + C$\n\n6. Solve the system of equations:\nFrom $0 = B + C$, we get $C = -B$.\nSubstitute $C = -B$ into $4 = 2A + C$:\n$$4 = 2A - B$$\nFrom $1 = A + B$, we get $A = 1 - B$.\nSubstitute $A = 1 - B$ into $4 = 2A - B$:\n$$4 = 2(1 - B) - B = 2 - 2B - B = 2 - 3B$$\nRearranged:\n$$4 - 2 = -3B \Rightarrow 2 = -3B \Rightarrow B = -\frac{2}{3}$$\n\n7. Find $A$ and $C$:\n$$A = 1 - B = 1 - \left(-\frac{2}{3}\right) = 1 + \frac{2}{3} = \frac{5}{3}$$\n$$C = -B = -\left(-\frac{2}{3}\right) = \frac{2}{3}$$\n\n8. Write the partial fraction decomposition:\n$$\frac{x^2 + 4}{(x+1)(x^2 + 2)} = \frac{5/3}{x+1} + \frac{-\frac{2}{3}x + \frac{2}{3}}{x^2 + 2} = \frac{5}{3(x+1)} + \frac{-2x + 2}{3(x^2 + 2)}$$\n\nFinal answer:\n$$\boxed{\frac{x^2 + 4}{(x+1)(x^2 + 2)} = \frac{5}{3(x+1)} + \frac{-2x + 2}{3(x^2 + 2)}}$$