Resolve `(x^(3)+x-1)/((x^(2)+1)(x^(2)+2x+3))` into partial fractions.

Answer

To resolve the expression \(\frac{x^3 + x - 1}{(x^2 + 1)(x^2 + 2x + 3)}\) into partial fractions, we follow these steps:

### Step 1: Set Up the Partial Fraction Decomposition
We start by expressing the given rational function as a sum of partial fractions:
\[ \frac{x^3 + x - 1}{(x^2 + 1)(x^2 + 2x + 3)} = \frac{ax + b}{x^2 + 1} + \frac{cx + d}{x^2 + 2x + 3} \]
Here, \(a\), \(b\), \(c\), and \(d\) are constants that we need to determine.

### Step 2: Combine the Right Side
Next, we combine the right-hand side into a single fraction:
\[ \frac{ax + b}{x^2 + 1} + \frac{cx + d}{x^2 + 2x + 3} = \frac{(ax + b)(x^2 + 2x + 3) + (cx + d)(x^2 + 1)}{(x^2 + 1)(x^2 + 2x + 3)} \]

### Step 3: Expand the Numerator
Now, we expand the numerator:
\[ (ax + b)(x^2 + 2x + 3) = ax^3 + 2ax^2 + 3ax + bx^2 + 2bx + 3b \]
\[ (cx + d)(x^2 + 1) = cx^3 + dx^2 + cx + d \]
Combining these gives:
\[ (ax^3 + cx^3) + (2ax^2 + bx^2 + dx^2) + (3ax + 2bx + cx) + (3b + d) \]
This simplifies to:
\[ (a + c)x^3 + (2a + b + d)x^2 + (3a + 2b + c)x + (3b + d) \]

### Step 4: Set Up the Equation
Now we equate the coefficients of the numerator to those of the original polynomial \(x^3 + x - 1\):
1. Coefficient of \(x^3\): \(a + c = 1\)
2. Coefficient of \(x^2\): \(2a + b + d = 0\)
3. Coefficient of \(x\): \(3a + 2b + c = 1\)
4. Constant term: \(3b + d = -1\)

### Step 5: Solve the System of Equations
We now have a system of four equations:
1. \(a + c = 1\) (Equation 1)
2. \(2a + b + d = 0\) (Equation 2)
3. \(3a + 2b + c = 1\) (Equation 3)
4. \(3b + d = -1\) (Equation 4)

From Equation 1, we can express \(c\) in terms of \(a\):
\[ c = 1 - a \]

Substituting \(c\) into Equation 3:
\[ 3a + 2b + (1 - a) = 1 \implies 2a + 2b = 0 \implies a + b = 0 \implies b = -a \]

Now substitute \(b = -a\) into Equation 2:
\[ 2a - a + d = 0 \implies a + d = 0 \implies d = -a \]

Now substitute \(b = -a\) and \(d = -a\) into Equation 4:
\[ 3(-a) + (-a) = -1 \implies -4a = -1 \implies a = \frac{1}{4} \]

### Step 6: Find Values of \(b\), \(c\), and \(d\)
Now we can find \(b\), \(c\), and \(d\):
\[ b = -a = -\frac{1}{4} \]
\[ c = 1 - a = 1 - \frac{1}{4} = \frac{3}{4} \]
\[ d = -a = -\frac{1}{4} \]

### Step 7: Write the Final Answer
Now we substitute \(a\), \(b\), \(c\), and \(d\) back into the partial fractions:
\[ \frac{x^3 + x - 1}{(x^2 + 1)(x^2 + 2x + 3)} = \frac{\frac{1}{4}x - \frac{1}{4}}{x^2 + 1} + \frac{\frac{3}{4}x - \frac{1}{4}}{x^2 + 2x + 3} \]

### Final Result
Thus, the partial fraction decomposition is:
\[ \frac{x^3 + x - 1}{(x^2 + 1)(x^2 + 2x + 3)} = \frac{\frac{1}{4}(x - 1)}{x^2 + 1} + \frac{\frac{3}{4}x - \frac{1}{4}}{x^2 + 2x + 3} \]