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

Answer

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

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

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

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

### Step 2: Clear the denominators
Next, we multiply both sides by the common denominator \((x + 1)(x - 1)^2\) to eliminate the fractions:

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

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

1. Expand \(A(x - 1)^2\):
\[ A(x^2 - 2x + 1) = Ax^2 - 2Ax + A \]

2. Expand \(B(x + 1)(x - 1)\):
\[ B(x^2 - 1) = Bx^2 - B \]

3. Expand \(C(x + 1)\):
\[ Cx + C \]

Combining these, we have:

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

### Step 4: Compare coefficients
Now we equate the coefficients from both sides:

- For \(x^2\): \(A + B = 1\) (1)
- For \(x\): \(-2A + C = -1\) (2)
- For the constant term: \(A - B + C = 1\) (3)

### Step 5: Solve the system of equations
We now have a system of three equations:

1. \(A + B = 1\)
2. \(-2A + C = -1\)
3. \(A - B + C = 1\)

From equation (1), we can express \(B\) in terms of \(A\):
\[ B = 1 - A \]

Substituting \(B\) into equation (3):
\[ A - (1 - A) + C = 1 \implies 2A - 1 + C = 1 \implies 2A + C = 2 \implies C = 2 - 2A \quad (4) \]

Now substitute \(C\) from equation (4) into equation (2):
\[ -2A + (2 - 2A) = -1 \implies -4A + 2 = -1 \implies -4A = -3 \implies A = \frac{3}{4} \]

Now substitute \(A\) back to find \(B\) and \(C\):
\[ B = 1 - \frac{3}{4} = \frac{1}{4} \]
\[ C = 2 - 2 \cdot \frac{3}{4} = 2 - \frac{3}{2} = \frac{1}{2} \]

### Step 6: Write the final partial fraction decomposition
Now we can substitute \(A\), \(B\), and \(C\) back into our partial fraction expression:

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

### Final Answer
Thus, the partial fraction decomposition is:

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