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The problem in the image asks you to decompose the given rational expression into partial fractions. The expression is:

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

### Step-by-step decomposition:

1. **Identify the form of the partial fractions**: 
   The denominator consists of two distinct factors:
   - \( x + 1 \) is a linear factor.
   - \( x^2 - x + 3 \) is a quadratic factor that doesn't factor further.

   So, the partial fraction decomposition will have the form:
   \[
   \frac{3x^2 - x + 1}{(x+1)(x^2 - x + 3)} = \frac{A}{x+1} + \frac{Bx + C}{x^2 - x + 3}
   \]
   where \( A \), \( B \), and \( C \) are constants to be determined.

2. **Clear the denominators** by multiplying both sides of the equation by \( (x+1)(x^2 - x + 3) \):
   \[
   3x^2 - x + 1 = A(x^2 - x + 3) + (Bx + C)(x + 1)
   \]

3. **Expand both sides**:
   - Expand the right-hand side:
     \[
     A(x^2 - x + 3) = A x^2 - A x + 3A
     \]
     \[
     (Bx + C)(x + 1) = Bx^2 + Bx + Cx + C = Bx^2 + (B + C)x + C
     \]
   - Combine everything:
     \[
     3x^2 - x + 1 = (A + B)x^2 + (-A + B + C)x + (3A + C)
     \]

4. **Set up a system of equations** by comparing the coefficients of like powers of \( x \):
   - Coefficient of \( x^2 \): \( A + B = 3 \)
   - Coefficient of \( x \): \( -A + B + C = -1 \)
   - Constant term: \( 3A + C = 1 \)

5. **Solve the system of equations**:
   - From \( A + B = 3 \), solve for \( B \): \( B = 3 - A \).
   - Substitute \( B = 3 - A \) into the other equations:
     \[
     -A + (3 - A) + C = -1 \quad \Rightarrow \quad -2A + 3 + C = -1 \quad \Rightarrow \quad -2A + C = -4
     \]
     \[
     3A + C = 1
     \]

     Now solve the system:
     - From \( -2A + C = -4 \), we get \( C = 2A - 4 \).
     - Substitute \( C = 2A - 4 \) into \( 3A + C = 1 \):
       \[
       3A + (2A - 4) = 1 \quad \Rightarrow \quad 5A - 4 = 1 \quad \Rightarrow \quad 5A = 5 \quad \Rightarrow \quad A = 1
       \]
     - Now, substitute \( A = 1 \) into \( B = 3 - A \): 
       \[
       B = 3 - 1 = 2
       \]
     - Finally, substitute \( A = 1 \) into \( C = 2A - 4 \):
       \[
       C = 2(1) - 4 = -2
       \]

6. **Write the final partial fraction decomposition**:
   \[
   \frac{3x^2 - x + 1}{(x+1)(x^2 - x + 3)} = \frac{1}{x+1} + \frac{2x - 2}{x^2 - x + 3}
   \]

This is the decomposed form of the given rational expression.

Would you like further clarification on any step? Here are some related questions to deepen your understanding:

1. What is the purpose of partial fraction decomposition in calculus?
2. How can you solve a system of equations using substitution or elimination?
3. Why is the quadratic factor \(x^2 - x + 3\) not factorable?
4. How would the decomposition change if the denominator had repeated linear factors?
5. Can partial fraction decomposition be applied to improper fractions?

**Tip:** Always check the degree of the numerator and denominator to determine whether you first need to perform polynomial long division before decomposing into partial fractions.

Decompose into partial fractions: (3x^2 - x + 1) / [(x + 1)(x^2 - x + 3)]

Learn how to decompose the rational expression (3x^2 - x + 1) / [(x+1)(x^2 - x + 3)] into partial fractions. This step-by-step guide covers the necessary algebraic techniques and systems of equations to find constants A, B, and C for the decomposition.

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