Resolve `(1)/(x^(2)(x+1))` into partial fractions.

Answer

To resolve the expression \( \frac{1}{x^2(x+1)} \) into partial fractions, we follow these steps:

### Step 1: Set up the partial fractions
We start by expressing \( \frac{1}{x^2(x+1)} \) in terms of its partial fractions. Since the denominator has a repeated linear factor \( x^2 \) and a simple linear factor \( (x+1) \), we can write:

\[ \frac{1}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} \]

where \( A \), \( B \), and \( C \) are constants that we need to determine.

### Step 2: Clear the denominators
To eliminate the denominators, we multiply both sides of the equation by \( x^2(x+1) \):

\[ 1 = A(x)(x+1) + B(x+1) + C(x^2) \]

### Step 3: Expand the right-hand side
Now, we expand the right-hand side:

\[ 1 = A(x^2 + x) + B(x + 1) + C(x^2) \]
\[ 1 = (A + C)x^2 + (A + B)x + B \]

### Step 4: Equate coefficients
Next, we equate the coefficients of \( x^2 \), \( x \), and the constant term from both sides of the equation. Since the left side has no \( x \) terms, we have:

1. Coefficient of \( x^2 \): \( A + C = 0 \)
2. Coefficient of \( x \): \( A + B = 0 \)
3. Constant term: \( B = 1 \)

### Step 5: Solve the system of equations
From the third equation, we have:

\[ B = 1 \]

Substituting \( B \) into the second equation:

\[ A + 1 = 0 \implies A = -1 \]

Now, substituting \( A \) into the first equation:

\[ -1 + C = 0 \implies C = 1 \]

### Step 6: Write the partial fraction decomposition
Now that we have the values of \( A \), \( B \), and \( C \):

\[ A = -1, \quad B = 1, \quad C = 1 \]

Thus, we can write the partial fraction decomposition as:

\[ \frac{1}{x^2(x+1)} = \frac{-1}{x} + \frac{1}{x^2} + \frac{1}{x+1} \]

### Final Result
The final result of the partial fraction decomposition is:

\[ \frac{1}{x^2(x+1)} = \frac{-1}{x} + \frac{1}{x^2} + \frac{1}{x+1} \]

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