The problem asks to perform partial fraction decomposition on the rational function $\frac{x+4}{(x+1)(x-2)^2}$.

1. Problem Description

The problem asks to perform partial fraction decomposition on the rational function $\frac{x+4}{(x+1)(x-2)^2}$.

2. Solution Steps

We want to decompose the given rational function into partial fractions. Since the denominator is $(x+1)(x-2)^2$, we write the decomposition as

$$\frac{x+4}{(x+1)(x-2)^2} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}$$

Multiplying both sides by $(x+1)(x-2)^2$, we get

$$x+4 = A(x-2)^2 + B(x+1)(x-2) + C(x+1)$$

Let $x = -1$. Then

$$-1+4 = A(-1-2)^2 + B(-1+1)(-1-2) + C(-1+1)$$

$$3 = A(-3)^2 + 0 + 0$$

$$3 = 9A$$

$$A = \frac{3}{9} = \frac{1}{3}$$

Let $x = 2$. Then

$$2+4 = A(2-2)^2 + B(2+1)(2-2) + C(2+1)$$

$$6 = 0 + 0 + 3C$$

$$3C = 6$$

$$C = \frac{6}{3} = 2$$

Now, we can pick any other value for $x$ to solve for $B$. Let's choose $x = 0$.

$$0+4 = A(0-2)^2 + B(0+1)(0-2) + C(0+1)$$

$$4 = 4A - 2B + C$$

Substitute $A = \frac{1}{3}$ and $C = 2$:

$$4 = 4\left(\frac{1}{3}\right) - 2B + 2$$

$$4 = \frac{4}{3} - 2B + 2$$

$$4 - 2 - \frac{4}{3} = -2B$$

$$2 - \frac{4}{3} = -2B$$

$$\frac{6-4}{3} = -2B$$

$$\frac{2}{3} = -2B$$

$$B = \frac{2/3}{-2} = -\frac{1}{3}$$

Thus, the partial fraction decomposition is

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

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

3. Final Answer

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