Question

Partial fractions(types and examples)

Answer 1

Partial fraction decomposition is a method used to break down rational functions into simpler fractions, making them easier to integrate or work with in other mathematical contexts. Here's a brief overview of the types of partial fractions and some examples.

### Types of Partial Fractions

1. **Linear Factors**: When the denominator can be factored into distinct linear factors (e.g., \( (x-a)(x-b) \)).
- **Example**: \( \frac{1}{(x-1)(x-2)} \)

2. **Repeated Linear Factors**: When the denominator has factors that are repeated (e.g., \( (x-a)^n \)).
- **Example**: \( \frac{1}{(x-1)^2} \)

3. **Irreducible Quadratic Factors**: When the denominator contains irreducible quadratic factors (e.g., \( x^2 + bx + c \)).
- **Example**: \( \frac{1}{(x^2+1)(x-2)} \)

4. **Repeated Irreducible Quadratic Factors**: Similar to irreducible quadratic factors but repeated (e.g., \( (x^2 + bx + c)^n \)).
- **Example**: \( \frac{1}{(x^2+1)^2} \)

### Example of Each Type

#### 1. Distinct Linear Factors
Consider the function:
\[
\frac{4}{(x-1)(x+2)}
\]
**Decomposition:**
\[
\frac{4}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}
\]
To find \( A \) and \( B \), multiply through by the denominator:
\[
4 = A(x+2) + B(x-1)
\]
By substituting suitable values or solving for coefficients, we can find:
Let \( x = 1 \): \( 4 = A(1 + 2) + B(0) \) → \( A = \frac{4}{3} \)
Let \( x = -2 \): \( 4 = A(0) + B(-2 - 1) \) → \( B = -\frac{4}{3} \)
Final decomposition:
\[
\frac{4}{(x-1)(x+2)} = \frac{\frac{4}{3}}{x-1} - \frac{\frac{4}{3}}{x+2}
\]

#### 2. Repeated Linear Factors
For:
\[
\frac{3}{(x-1)^2}
\]
**Decomposition:**
\[
\frac{3}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2}
\]
Multiply through:
\[
3 = A(x-1) + B
\]
Let \( x = 1 \): \( 3 = B \rightarrow B = 3 \)
Let \( B = 0 \): \( 3 = 0(x - 1) + 3 \)
Thus, the final result is:
\[
\frac{3}{(x-1)^2} = \frac{3}{(x-1)^2}
\]

#### 3. Irreducible Quadratic Factors
For:
\[
\frac{5}{(x^2 + 1)(x-2)}
\]
**Decomposition:**
\[
\frac{5}{(x^2+1)(x-2)} = \frac{Ax + B}{x^2 + 1} + \frac{C}{x - 2}
\]
Multiply through:
\[
5 = (Ax + B)(x - 2) + C(x^2 + 1)
\]
Expanding and collecting like terms gives a system of equations to solve for \( A, B, C \).

#### 4. Repeated Irreducible Quadratic Factors
For:
\[
\frac{6}{(x^2 + 1)^2}
\]
**Decomposition:**
\[
\frac{6}{(x^2+1)^2} = \frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{(x^2 + 1)^2}
\]
Multiply through and expand, then equate coefficients to solve for \( A, B, C, D \).

### Conclusion
Partial fraction decomposition is a powerful technique in algebra and calculus, allowing for the simplification of complex rational expressions. Each type has a systematic approach for decomposition. The examples provided illustrate the primary methodologies used for these different forms.