Question
Comprehensive explanation and examples under all types if partial fractions
Answers
GPT-4o mini
Partial fraction decomposition is a technique used to break down rational functions into simpler fractions that can be more easily integrated or manipulated algebraically. When dealing with a rational function \( \frac{P(x)}{Q(x)} \), where \(P(x)\) and \(Q(x)\) are polynomials, we can express it as a sum of simpler fractions. The method of partial fractions applies differently depending on the factors of the denominator \(Q(x)\).
Here's a comprehensive explanation of the different types of partial fractions and their corresponding examples:
### 1. Distinct Linear Factors
If the denominator \(Q(x)\) can be factored into distinct linear factors, we can express the partial fractions in the form:
\[
\frac{A_1}{(x - r_1)} + \frac{A_2}{(x - r_2)} + ... + \frac{A_n}{(x - r_n)}
\]
Where \( r_1, r_2, \ldots, r_n \) are the distinct roots of \(Q(x)\) and \(A_1, A_2, \ldots, A_n\) are constants that need to be determined.
**Example:**
\[
\frac{2x + 3}{(x - 1)(x + 2)}
\]
We want to write this as:
\[
\frac{A}{x - 1} + \frac{B}{x + 2}
\]
Multiply through by the denominator \((x - 1)(x + 2)\):
\[
2x + 3 = A(x + 2) + B(x - 1)
\]
Expanding the right-hand side:
\[
2x + 3 = Ax + 2A + Bx - B = (A + B)x + (2A - B)
\]
Now we equate coefficients:
1. \( A + B = 2 \)
2. \( 2A - B = 3 \)
Solving this system, we find \( A = 2 \) and \( B = 0 \). Thus:
\[
\frac{2x + 3}{(x - 1)(x + 2)} = \frac{2}{x - 1} + \frac{0}{x + 2} = \frac{2}{x - 1}
\]
### 2. Repeated Linear Factors
If \(Q(x)\) contains repeated linear factors, the form is slightly more complex:
\[
\frac{A_1}{(x - r)^1} + \frac{A_2}{(x - r)^2} + ... + \frac{A_n}{(x - r)^n}
\]
Where \(r\) is the repeated root.
**Example:**
\[
\frac{3x + 5}{(x - 1)^2}
\]
We write this as:
\[
\frac{A}{x - 1} + \frac{B}{(x - 1)^2}
\]
Multiplying through by \((x - 1)^2\):
\[
3x + 5 = A(x - 1) + B
\]
Expanding:
\[
3x + 5 = Ax - A + B = Ax + (B - A)
\]
Equating coefficients yields:
1. \( A = 3 \)
2. \( B - A = 5 \Rightarrow B - 3 = 5 \Rightarrow B = 8 \)
So,
\[
\frac{3x + 5}{(x - 1)^2} = \frac{3}{x - 1} + \frac{8}{(x - 1)^2}
\]
### 3. Irreducible Quadratic Factors
If \(Q(x)\) contains irreducible quadratic factors of the form \( ax^2 + bx + c \) (which cannot be factored further over the reals), we use the following form:
\[
\frac{Ax + B}{ax^2 + bx + c}
\]
for each irreducible quadratic factor.
**Example:**
\[
\frac{x^2 + 3}{(x^2 + 1)(x - 2)}
\]
We write this as:
\[
\frac{Ax + B}{x^2 + 1} + \frac{C}{x - 2}
\]
Multiplying through by the denominator \((x^2 + 1)(x - 2)\):
\[
x^2 + 3 = (Ax + B)(x - 2) + C(x^2 + 1)
\]
Expanding:
\[
x^2 + 3 = Ax^2 - 2Ax + Bx - 2B + Cx^2 + C
\]
\[
= (A + C)x^2 + (-2A + B)x + (C - 2B)
\]
Equating coefficients:
1. \(A + C = 1\)
2. \(-2A + B = 0\)
3. \(C - 2B = 3\)
Solving this system gives \(A = 1\), \(B = 2\), and \(C = 0\). Thus,
\[
\frac{x^2 + 3}{(x^2 + 1)(x - 2)} = \frac{1x + 2}{x^2 + 1} + \frac{0}{x - 2} = \frac{x + 2}{x^2 + 1}
\]
### Summary
To summarize, the steps for using partial fractions are as follows:
1. **Factor the Denominator** completely into distinct linear, repeated linear, and irreducible quadratic factors.
2. **Set up the partial fraction decomposition** using the appropriate forms.
3. **Clear the denominators** by multiplying through by \(Q(x)\).
4. **Expand and equate coefficients** for corresponding powers of \(x\) to solve for constants \(A, B, C, \ldots\).
5. **Substitute back** into the original equation to obtain the partial fraction decomposition.
This technique is fundamental in calculus, particularly when performing integration of rational functions. Each type of factor requires a slightly different approach, but the underlying principles remain the same.
Here's a comprehensive explanation of the different types of partial fractions and their corresponding examples:
### 1. Distinct Linear Factors
If the denominator \(Q(x)\) can be factored into distinct linear factors, we can express the partial fractions in the form:
\[
\frac{A_1}{(x - r_1)} + \frac{A_2}{(x - r_2)} + ... + \frac{A_n}{(x - r_n)}
\]
Where \( r_1, r_2, \ldots, r_n \) are the distinct roots of \(Q(x)\) and \(A_1, A_2, \ldots, A_n\) are constants that need to be determined.
**Example:**
\[
\frac{2x + 3}{(x - 1)(x + 2)}
\]
We want to write this as:
\[
\frac{A}{x - 1} + \frac{B}{x + 2}
\]
Multiply through by the denominator \((x - 1)(x + 2)\):
\[
2x + 3 = A(x + 2) + B(x - 1)
\]
Expanding the right-hand side:
\[
2x + 3 = Ax + 2A + Bx - B = (A + B)x + (2A - B)
\]
Now we equate coefficients:
1. \( A + B = 2 \)
2. \( 2A - B = 3 \)
Solving this system, we find \( A = 2 \) and \( B = 0 \). Thus:
\[
\frac{2x + 3}{(x - 1)(x + 2)} = \frac{2}{x - 1} + \frac{0}{x + 2} = \frac{2}{x - 1}
\]
### 2. Repeated Linear Factors
If \(Q(x)\) contains repeated linear factors, the form is slightly more complex:
\[
\frac{A_1}{(x - r)^1} + \frac{A_2}{(x - r)^2} + ... + \frac{A_n}{(x - r)^n}
\]
Where \(r\) is the repeated root.
**Example:**
\[
\frac{3x + 5}{(x - 1)^2}
\]
We write this as:
\[
\frac{A}{x - 1} + \frac{B}{(x - 1)^2}
\]
Multiplying through by \((x - 1)^2\):
\[
3x + 5 = A(x - 1) + B
\]
Expanding:
\[
3x + 5 = Ax - A + B = Ax + (B - A)
\]
Equating coefficients yields:
1. \( A = 3 \)
2. \( B - A = 5 \Rightarrow B - 3 = 5 \Rightarrow B = 8 \)
So,
\[
\frac{3x + 5}{(x - 1)^2} = \frac{3}{x - 1} + \frac{8}{(x - 1)^2}
\]
### 3. Irreducible Quadratic Factors
If \(Q(x)\) contains irreducible quadratic factors of the form \( ax^2 + bx + c \) (which cannot be factored further over the reals), we use the following form:
\[
\frac{Ax + B}{ax^2 + bx + c}
\]
for each irreducible quadratic factor.
**Example:**
\[
\frac{x^2 + 3}{(x^2 + 1)(x - 2)}
\]
We write this as:
\[
\frac{Ax + B}{x^2 + 1} + \frac{C}{x - 2}
\]
Multiplying through by the denominator \((x^2 + 1)(x - 2)\):
\[
x^2 + 3 = (Ax + B)(x - 2) + C(x^2 + 1)
\]
Expanding:
\[
x^2 + 3 = Ax^2 - 2Ax + Bx - 2B + Cx^2 + C
\]
\[
= (A + C)x^2 + (-2A + B)x + (C - 2B)
\]
Equating coefficients:
1. \(A + C = 1\)
2. \(-2A + B = 0\)
3. \(C - 2B = 3\)
Solving this system gives \(A = 1\), \(B = 2\), and \(C = 0\). Thus,
\[
\frac{x^2 + 3}{(x^2 + 1)(x - 2)} = \frac{1x + 2}{x^2 + 1} + \frac{0}{x - 2} = \frac{x + 2}{x^2 + 1}
\]
### Summary
To summarize, the steps for using partial fractions are as follows:
1. **Factor the Denominator** completely into distinct linear, repeated linear, and irreducible quadratic factors.
2. **Set up the partial fraction decomposition** using the appropriate forms.
3. **Clear the denominators** by multiplying through by \(Q(x)\).
4. **Expand and equate coefficients** for corresponding powers of \(x\) to solve for constants \(A, B, C, \ldots\).
5. **Substitute back** into the original equation to obtain the partial fraction decomposition.
This technique is fundamental in calculus, particularly when performing integration of rational functions. Each type of factor requires a slightly different approach, but the underlying principles remain the same.