The problem asks us to decompose the rational function $\frac{x+4}{(x+1)(x-2)^2}$ into partial fractions.

1. Problem Description

The problem asks us to decompose the rational function $\frac{x+4}{(x+1)(x-2)^2}$ into partial fractions.

2. Solution Steps

Since the denominator has a linear factor $(x+1)$ and a repeated linear factor $(x-2)^2$, we can write the partial fraction 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:

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

Now we solve for $A$, $B$, and $C$.

Let $x = -1$:

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

$3 = A(-3)^2$

$3 = 9A$

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

Let $x = 2$:

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

$6 = 0 + 0 + 3C$

$3C = 6$

$C = 2$

Now we need to find $B$. We can substitute $A = \frac{1}{3}$ and $C = 2$ and choose any other value for $x$. Let $x=0$:

$0 + 4 = \frac{1}{3}(0-2)^2 + B(0+1)(0-2) + 2(0+1)$

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

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

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

$4 = \frac{10}{3} - 2B$

$12 = 10 - 6B$

$2 = -6B$

$B = -\frac{2}{6} = -\frac{1}{3}$

Therefore, we have $A = \frac{1}{3}$, $B = -\frac{1}{3}$, and $C = 2$.

The partial fraction decomposition is:

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

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

3. Final Answer

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