Determine The Partial Fraction Expansion For The Rational Function Below

Determining the Partial Fraction Expansion for Rational Functions: A Comprehensive Guide

Partial fraction decomposition is a crucial technique in calculus and other branches of mathematics, particularly when dealing with integration of rational functions. A rational function is simply a ratio of two polynomials, P(x)/Q(x), where P(x) is the numerator and Q(x) is the denominator. However, integrating such functions directly can be challenging. Partial fraction expansion transforms a complex rational function into a sum of simpler fractions, making integration significantly easier. This comprehensive guide will delve into the process, covering various scenarios and providing practical examples.

Understanding the Fundamentals

Before diving into the techniques, let's establish a solid foundation:

• Proper Rational Function: A proper rational function is one where the degree of the numerator polynomial is strictly less than the degree of the denominator polynomial (deg(P(x)) < deg(Q(x))). If this condition isn't met (it's an improper rational function), you'll need to perform polynomial long division first to obtain a proper rational function plus a polynomial. The polynomial part is easily integrated, leaving the proper rational function for partial fraction decomposition.

• Distinct Linear Factors: If the denominator Q(x) can be factored into distinct linear factors, Q(x) = (x - a)(x - b)(x - c)..., then the partial fraction decomposition takes the form:

  P(x)/Q(x) = A/(x - a) + B/(x - b) + C/(x - c) + ...

  where A, B, C, etc., are constants to be determined.

• Repeated Linear Factors: If Q(x) contains repeated linear factors, such as (x - a)^n, the partial fraction decomposition includes terms for each power of the repeated factor:

  P(x)/[(x - a)^n * ...] = A/(x - a) + B/(x - a)^2 + C/(x - a)^3 + ... + ...

• Irreducible Quadratic Factors: If Q(x) contains irreducible quadratic factors (factors that cannot be factored into real linear factors, such as x² + 1), the corresponding terms in the partial fraction decomposition will have the form:

  (Ax + B)/(x² + px + q)

  where A and B are constants to be determined.

• Repeated Irreducible Quadratic Factors: Similar to repeated linear factors, repeated irreducible quadratic factors will require terms for each power of the repeated factor. For example, for (x² + px + q)^n, you'll have terms like:

  (Ax + B)/(x² + px + q) + (Cx + D)/(x² + px + q)² + ...

Methods for Determining Constants

Several methods exist for determining the constants (A, B, C, etc.) in the partial fraction expansion. The most common are:

1. The Heaviside Cover-Up Method (for distinct linear factors):

This method provides a quick way to find the constants when the denominator consists solely of distinct linear factors. For each term, A/(x - a), you cover up the (x - a) factor in the denominator of the original rational function and substitute x = a into the remaining expression. The result is the constant A.

Example:

Let's find the partial fraction expansion of f(x) = (3x + 5)/[(x - 1)(x + 2)].

1. The distinct linear factors are (x - 1) and (x + 2). Therefore, the partial fraction form is:

   (3x + 5)/[(x - 1)(x + 2)] = A/(x - 1) + B/(x + 2)

2. To find A, cover up (x - 1) and substitute x = 1:

   A = (3(1) + 5)/(1 + 2) = 8/3

3. To find B, cover up (x + 2) and substitute x = -2:

   B = (3(-2) + 5)/(-2 - 1) = -1/-3 = 1/3

4. Therefore, the partial fraction expansion is:

   (3x + 5)/[(x - 1)(x + 2)] = 8/3(x - 1) + 1/3(x + 2)

2. The Method of Equating Coefficients:

This is a more general method applicable to all types of partial fraction decompositions. After expressing the rational function in its partial fraction form, you multiply both sides by the common denominator. This will result in an equation involving polynomials. You then equate the coefficients of corresponding powers of x on both sides to obtain a system of linear equations. Solving this system gives the values of the constants.

Example:

Let's use this method for the same example: f(x) = (3x + 5)/[(x - 1)(x + 2)].

1. We have: (3x + 5)/[(x - 1)(x + 2)] = A/(x - 1) + B/(x + 2)

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

   3x + 5 = A(x + 2) + B(x - 1)

3. Expand and equate coefficients:

   3x + 5 = (A + B)x + (2A - B)

   Equating coefficients of x: A + B = 3 Equating constant terms: 2A - B = 5

4. Solve the system of equations (e.g., using substitution or elimination): Adding the two equations gives 3A = 8, so A = 8/3. Substituting this into A + B = 3 gives B = 1/3.

5. This confirms the same result as the Heaviside method.

3. Using a System of Linear Equations (for more complex cases):

For more complicated scenarios with repeated factors or irreducible quadratic factors, the method of equating coefficients is often the most straightforward. You'll end up with a system of linear equations that can be solved using techniques like Gaussian elimination, matrix methods, or substitution.

Advanced Scenarios and Examples

Let's consider some more complex examples to illustrate the application of these techniques:

Example 1: Repeated Linear Factor

Find the partial fraction decomposition of: f(x) = (x² + 2x + 3)/[(x - 1)²(x + 2)]

The partial fraction form is:

(x² + 2x + 3)/[(x - 1)²(x + 2)] = A/(x - 1) + B/(x - 1)² + C/(x + 2)

Multiplying by the common denominator and equating coefficients leads to a system of three equations in three unknowns (A, B, C). Solving this system will yield the values for A, B, and C.

Example 2: Irreducible Quadratic Factor

Find the partial fraction decomposition of: f(x) = (2x³ + x² + 3x + 1)/[(x² + 1)(x + 1)]

Since x² + 1 is an irreducible quadratic factor, the partial fraction form will be:

(2x³ + x² + 3x + 1)/[(x² + 1)(x + 1)] = (Ax + B)/(x² + 1) + C/(x + 1)

Again, multiplying by the common denominator and equating coefficients leads to a system of linear equations to solve for A, B, and C.

Example 3: Improper Rational Function

Consider f(x) = (x³ + 2x² + 3x + 4)/(x² + 1).

Because this is an improper rational function (the degree of the numerator is greater than the degree of the denominator), we first perform polynomial long division:

x³ + 2x² + 3x + 4 = (x + 2)(x² + 1) + x + 2

Therefore: f(x) = x + 2 + (x + 2)/(x² + 1)

Now, we only need to decompose the remaining proper rational function (x + 2)/(x² + 1). This is a relatively simple case involving an irreducible quadratic factor.

Conclusion

Partial fraction decomposition is a powerful tool for simplifying complex rational functions, making them significantly easier to integrate and manipulate in various mathematical contexts. While the Heaviside cover-up method offers a shortcut for distinct linear factors, the method of equating coefficients provides a more general approach applicable to all cases. Mastery of these techniques is essential for anyone working with calculus and related fields. Remember to always check your solutions by combining the partial fractions to ensure they reconstruct the original rational function. Practice with a wide range of examples is key to developing proficiency in this important technique.