Determine The Partial Fraction Expansion For The Rational Function Below

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Decomposing rational functions into partial fractions is a fundamental technique in calculus, differential equations, and other areas of mathematics and engineering. It allows us to break down complex rational expressions into simpler components that are easier to integrate, manipulate, or analyze. This process is particularly valuable when dealing with integrals involving rational functions or when solving differential equations using Laplace transforms. The core idea is to express a rational function as a sum of fractions with simpler denominators derived from the factors of the original denominator.

Why Partial Fraction Decomposition?

Imagine trying to integrate a complicated rational function directly. It can be a daunting task. Partial fraction decomposition simplifies this process by allowing us to express the complex rational function as a sum of simpler fractions, each of which is much easier to integrate. Here's why it's so useful:

• Integration: Makes integration of rational functions significantly easier.
• Laplace Transforms: Used in finding inverse Laplace transforms to solve differential equations.
• Series Expansions: Aids in finding series representations of rational functions.
• Simplification: Simplifies algebraic manipulations and analysis of rational functions.

Prerequisites

Before diving into the methods, let's ensure we have a solid foundation. You'll need to be comfortable with:

• Polynomials: Understanding polynomial arithmetic, factoring, and long division.
• Rational Functions: Knowing how to manipulate and simplify rational expressions.
• Linear Equations: Solving systems of linear equations, which often arises in determining the coefficients of the partial fractions.
• Calculus Basics: A basic understanding of integration, as the main goal is to simplify integrals.

General Form of Partial Fraction Decomposition

A rational function is a function of the form P(x)/Q(x), where P(x) and Q(x) are polynomials. The method of partial fraction decomposition applies when the degree of P(x) is less than the degree of Q(x). If this is not the case, we first perform polynomial long division to obtain:

P(x)/Q(x) = S(x) + R(x)/Q(x)

where S(x) is a polynomial (the quotient) and R(x) is a polynomial with a degree less than that of Q(x) (the remainder). We then focus on decomposing R(x)/Q(x).

The decomposition depends on the factors of Q(x):

1. Linear Factors: For each non-repeated linear factor (ax + b) in Q(x), there is a term A/(ax + b) in the decomposition.
2. Repeated Linear Factors: For each repeated linear factor (ax + b)^n in Q(x), there are terms A1/(ax + b) + A2/(ax + b)^2 + ... + An/(ax + b)^n in the decomposition.
3. Irreducible Quadratic Factors: For each non-repeated irreducible quadratic factor (ax^2 + bx + c) in Q(x), there is a term (Ax + B)/(ax^2 + bx + c) in the decomposition.
4. Repeated Irreducible Quadratic Factors: For each repeated irreducible quadratic factor (ax^2 + bx + c)^n in Q(x), there are terms (A1x + B1)/(ax^2 + bx + c) + (A2x + B2)/(ax^2 + bx + c)^2 + ... + (Anx + Bn)/(ax^2 + bx + c)^n in the decomposition.

Step-by-Step Guide to Partial Fraction Decomposition

Let's break down the process into manageable steps:

Step 1: Check if the Rational Function is Proper

A rational function P(x)/Q(x) is proper if the degree of P(x) is less than the degree of Q(x). If it's not, perform polynomial long division.

Example:

Consider the rational function (x^3 + 2x^2 + x + 1) / (x^2 + x - 2). The degree of the numerator (3) is greater than the degree of the denominator (2), so it's an improper fraction. We perform long division:

        x + 1
x^2+x-2 | x^3 + 2x^2 + x + 1
          x^3 + x^2 - 2x
          ----------------
                x^2 + 3x + 1
                x^2 + x - 2
                ------------
                      2x + 3

So, (x^3 + 2x^2 + x + 1) / (x^2 + x - 2) = (x + 1) + (2x + 3) / (x^2 + x - 2). Now we focus on decomposing the proper fraction (2x + 3) / (x^2 + x - 2).

Step 2: Factor the Denominator Q(x)

Factor the denominator Q(x) into linear and irreducible quadratic factors. This is a crucial step, and the complexity of the decomposition heavily depends on how easily you can factor the denominator.

Example:

For the fraction (2x + 3) / (x^2 + x - 2), factor the denominator:

x^2 + x - 2 = (x + 2)(x - 1)

Step 3: Set Up the Partial Fraction Decomposition

Based on the factors of Q(x), set up the form of the partial fraction decomposition.

• Linear Factors: For each factor (x - a), include a term A/(x - a).
• Repeated Linear Factors: For each factor (x - a)^n, include terms A1/(x - a) + A2/(x - a)^2 + ... + An/(x - a)^n.
• Irreducible Quadratic Factors: For each factor (ax^2 + bx + c), include a term (Ax + B)/(ax^2 + bx + c).

Example:

Since x^2 + x - 2 = (x + 2)(x - 1), the partial fraction decomposition of (2x + 3) / (x^2 + x - 2) will have the form:

(2x + 3) / ((x + 2)(x - 1)) = A/(x + 2) + B/(x - 1)

Step 4: Determine the Unknown Coefficients

There are two primary methods to find the unknown coefficients (A, B, C, etc.):

• Method of Clearing Denominators: Multiply both sides of the equation by the original denominator Q(x) to clear the fractions. Then, either:
  • Substitute specific values of x that make the factors zero (Heaviside Cover-Up Method).
  • Equate coefficients of like powers of x on both sides of the equation and solve the resulting system of linear equations.
• Solving Systems of Linear Equations: After clearing denominators, equate coefficients of like powers of x and solve the resulting system of linear equations.

Example (Clearing Denominators and Substituting Values):

Starting with: (2x + 3) / ((x + 2)(x - 1)) = A/(x + 2) + B/(x - 1)

Multiply both sides by (x + 2)(x - 1):

2x + 3 = A(x - 1) + B(x + 2)

Now, substitute values of x to solve for A and B:

• Let x = 1: 2(1) + 3 = A(1 - 1) + B(1 + 2) => 5 = 3B => B = 5/3
• Let x = -2: 2(-2) + 3 = A(-2 - 1) + B(-2 + 2) => -1 = -3A => A = 1/3

Example (Equating Coefficients):

Starting with: 2x + 3 = A(x - 1) + B(x + 2)

Expand and group like terms:

2x + 3 = Ax - A + Bx + 2B

2x + 3 = (A + B)x + (-A + 2B)

Equate coefficients:

• Coefficient of x: A + B = 2
• Constant term: -A + 2B = 3

Solve the system of equations:

From the first equation, A = 2 - B. Substitute into the second equation:

-(2 - B) + 2B = 3 => -2 + B + 2B = 3 => 3B = 5 => B = 5/3

Then, A = 2 - 5/3 = 1/3

Step 5: Write the Partial Fraction Decomposition

Substitute the values of the coefficients back into the partial fraction decomposition.

Example:

Since A = 1/3 and B = 5/3, the partial fraction decomposition is:

(2x + 3) / (x^2 + x - 2) = (1/3)/(x + 2) + (5/3)/(x - 1)

Which can be written as:

(2x + 3) / (x^2 + x - 2) = 1/(3(x + 2)) + 5/(3(x - 1))

And going back to the original improper fraction:

(x^3 + 2x^2 + x + 1) / (x^2 + x - 2) = (x + 1) + 1/(3(x + 2)) + 5/(3(x - 1))

Examples

Example 1: Distinct Linear Factors

Decompose (5x - 4) / (x^2 - x - 2) into partial fractions.

1. Proper Fraction: Yes, degree of numerator (1) < degree of denominator (2).
2. Factor Denominator: x^2 - x - 2 = (x - 2)(x + 1)
3. Set Up Decomposition: (5x - 4) / ((x - 2)(x + 1)) = A/(x - 2) + B/(x + 1)
4. Find Coefficients:
   • Clear Denominators: 5x - 4 = A(x + 1) + B(x - 2)
   • Substitute x = 2: 5(2) - 4 = A(2 + 1) + B(2 - 2) => 6 = 3A => A = 2
   • Substitute x = -1: 5(-1) - 4 = A(-1 + 1) + B(-1 - 2) => -9 = -3B => B = 3
5. Write Decomposition: (5x - 4) / (x^2 - x - 2) = 2/(x - 2) + 3/(x + 1)

Example 2: Repeated Linear Factors

Decompose (x + 2) / (x(x - 1)^2) into partial fractions.

1. Proper Fraction: Yes.
2. Factor Denominator: x(x - 1)^2
3. Set Up Decomposition: (x + 2) / (x(x - 1)^2) = A/x + B/(x - 1) + C/(x - 1)^2
4. Find Coefficients:
   • Clear Denominators: x + 2 = A(x - 1)^2 + Bx(x - 1) + Cx
   • Substitute x = 0: 0 + 2 = A(0 - 1)^2 + B(0) + C(0) => 2 = A => A = 2
   • Substitute x = 1: 1 + 2 = A(1 - 1)^2 + B(1)(1 - 1) + C(1) => 3 = C => C = 3
   • Expand and Equate Coefficients: x + 2 = 2(x^2 - 2x + 1) + Bx(x - 1) + 3x x + 2 = 2x^2 - 4x + 2 + Bx^2 - Bx + 3x x + 2 = (2 + B)x^2 + (-4 - B + 3)x + 2
     • Coefficient of x^2: 0 = 2 + B => B = -2
5. Write Decomposition: (x + 2) / (x(x - 1)^2) = 2/x - 2/(x - 1) + 3/(x - 1)^2

Example 3: Irreducible Quadratic Factors

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

1. Proper Fraction: Yes.
2. Factor Denominator: x^3 + x = x(x^2 + 1)
3. Set Up Decomposition: (x^2 + 1) / (x(x^2 + 1)) = A/x + (Bx + C)/(x^2 + 1)
4. Find Coefficients:
   • Clear Denominators: x^2 + 1 = A(x^2 + 1) + (Bx + C)x
   • Substitute x = 0: 0 + 1 = A(0 + 1) + (B(0) + C)(0) => 1 = A => A = 1
   • Expand and Equate Coefficients: x^2 + 1 = 1(x^2 + 1) + Bx^2 + Cx x^2 + 1 = x^2 + 1 + Bx^2 + Cx x^2 + 1 = (1 + B)x^2 + Cx + 1
     • Coefficient of x^2: 1 = 1 + B => B = 0
     • Coefficient of x: 0 = C => C = 0
5. Write Decomposition: (x^2 + 1) / (x^3 + x) = 1/x + (0x + 0)/(x^2 + 1) = 1/x

Common Mistakes to Avoid

• Improper Fractions: Forgetting to perform long division when the degree of the numerator is greater than or equal to the degree of the denominator.
• Incorrect Decomposition Setup: Not including all necessary terms in the partial fraction decomposition, especially for repeated factors.
• Algebraic Errors: Making mistakes while clearing denominators, expanding terms, or solving systems of equations. Double-check your algebra!
• Incorrect Factoring: Failing to factor the denominator correctly. This is a critical step, so take your time and use appropriate factoring techniques.
• Not Checking Your Work: After finding the partial fraction decomposition, recombine the fractions to see if you get back the original rational function. This is a good way to catch errors.

Advanced Techniques and Considerations

• Heaviside Cover-Up Method: A quick way to find coefficients when dealing with distinct linear factors.
• Complex Factors: Sometimes, the denominator may have complex roots. The partial fraction decomposition can still be performed, but the coefficients and resulting fractions will be complex.
• Computer Algebra Systems (CAS): Software like Mathematica, Maple, or SymPy (Python library) can perform partial fraction decomposition, especially for complex rational functions.

Applications in Engineering and Physics

Partial fraction decomposition is not just a theoretical mathematical technique; it has many practical applications:

• Electrical Engineering: Analyzing circuits, especially when dealing with Laplace transforms to solve differential equations describing circuit behavior.
• Mechanical Engineering: Analyzing vibrations and control systems, where rational functions often arise in transfer functions.
• Chemical Engineering: Analyzing chemical reactions and transport phenomena, where rational functions can model concentrations and flow rates.
• Physics: Solving problems in electromagnetism and quantum mechanics, where rational functions can appear in Green's functions and scattering amplitudes.

Conclusion

Mastering partial fraction decomposition is a valuable skill for anyone working with calculus, differential equations, or engineering mathematics. By understanding the underlying principles and practicing the step-by-step methods, you can simplify complex rational functions and make them easier to analyze and manipulate. Remember to always check if the rational function is proper, factor the denominator carefully, set up the decomposition correctly, and double-check your algebra. With practice, you'll become proficient at this essential technique.