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

Answer

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

### Step 1: Set up the partial fraction decomposition
We want to express the given fraction as a sum of simpler fractions. Since the denominator consists of two quadratic factors, we can write:

\[ \frac{x^3 + x^2 + 1}{(x^2 + 2)(x^2 + 3)} = \frac{Ax + B}{x^2 + 2} + \frac{Cx + D}{x^2 + 3} \]

where \(A\), \(B\), \(C\), and \(D\) are constants to be determined.

### Step 2: Combine the right-hand side
To combine the fractions on the right-hand side, we find a common denominator:

\[ \frac{Ax + B}{x^2 + 2} + \frac{Cx + D}{x^2 + 3} = \frac{(Ax + B)(x^2 + 3) + (Cx + D)(x^2 + 2)}{(x^2 + 2)(x^2 + 3)} \]

### Step 3: Expand the numerator
Now, we expand the numerator:

\[ (Ax + B)(x^2 + 3) = Ax^3 + 3Ax + Bx^2 + 3B \]
\[ (Cx + D)(x^2 + 2) = Cx^3 + 2Cx + Dx^2 + 2D \]

Combining these gives:

\[ (A + C)x^3 + (B + D)x^2 + (3A + 2C)x + (3B + 2D) \]

### Step 4: Set up the equation
Now we can equate the numerators:

\[ x^3 + x^2 + 1 = (A + C)x^3 + (B + D)x^2 + (3A + 2C)x + (3B + 2D) \]

### Step 5: Compare coefficients
We can compare the coefficients of \(x^3\), \(x^2\), \(x\), and the constant term:

1. For \(x^3\): \(A + C = 1\) (Equation 1)
2. For \(x^2\): \(B + D = 1\) (Equation 2)
3. For \(x\): \(3A + 2C = 0\) (Equation 3)
4. For the constant: \(3B + 2D = 1\) (Equation 4)

### Step 6: Solve the equations
From Equation 1, we can express \(C\) in terms of \(A\):

\[ C = 1 - A \quad \text{(Substituting into Equation 3)} \]

Substituting \(C\) into Equation 3:

\[ 3A + 2(1 - A) = 0 \implies 3A + 2 - 2A = 0 \implies A + 2 = 0 \implies A = -2 \]

Now substituting \(A = -2\) back into Equation 1:

\[ -2 + C = 1 \implies C = 3 \]

Now we can find \(B\) and \(D\) using Equations 2 and 4. From Equation 2:

\[ B + D = 1 \quad \text{(Substituting into Equation 4)} \]

Substituting \(B\) into Equation 4:

\[ 3B + 2(1 - B) = 1 \implies 3B + 2 - 2B = 1 \implies B + 2 = 1 \implies B = -1 \]

Now substituting \(B = -1\) back into Equation 2:

\[ -1 + D = 1 \implies D = 2 \]

### Step 7: Write the final partial fraction decomposition
Now we have:

- \(A = -2\)
- \(B = -1\)
- \(C = 3\)
- \(D = 2\)

Thus, the partial fraction decomposition is:

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

### Final Answer
\[ \frac{x^3 + x^2 + 1}{(x^2 + 2)(x^2 + 3)} = \frac{-2x - 1}{x^2 + 2} + \frac{3x + 2}{x^2 + 3} \]