Partial Fractions

Understand the concept of partial fraction decomposition.

Use partial fraction decomposition with linear factors to integrate rational functions.

Use partial fraction decomposition with quadratic factors to integrate rational functions.

Objectives

Partial Fractions

Partial Fractions

The method of partial fractions is a procedure for decomposing a rational function into simpler rational functions to which you can apply the basic integration formulas.

To see the benefit of the method of partial fractions, consider the integral

Partial Fractions

Now, suppose you had observed that

Then you could evaluate the integral easily, as follows.

However, its use depends on the ability to factor the denominator, x2 – 5x + 6, and to find the partial fractions

Partial fraction decomposition

Partial Fractions

Linear Factors

Example 1 – Distinct Linear Factors

Write the partial fraction decomposition for

Solution:

Because x2 – 5x + 6 = (x – 3)(x – 2), you should include one partial fraction for each factor and write

where A and B are to be determined.

Multiplying this equation by the least common denominator (x – 3)(x – 2) yields the basic equation

1 = A(x – 2) + B(x – 3). Basic equation.

Example 1 – SolutionTo solve for A, let x = 3 and obtain

0x+1 = Ax+Bx-2A – 3B

Equating x0 terms

1 = -2A -3B (1)

and Equating x1 terms

0 = A+B

so -A=B (2)

Substitution of (2) into (1) yields

1=2B-3B

1=-1B

thus -1=B (3)

and A=1 (4)

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Example 1 – Solution

So, the decomposition is

Restating the original equation

and substituting A=1, B=-1 as shown at the beginning of this section, we get

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Quadratic Factors

Example 3 – Distinct Linear and Quadratic Factors

Find

Solution:

Because (x2 – x)(x2 + 4) = x(x – 1)(x2 + 4) you should include one partial fraction for each factor and write

Multiplying by the least common denominator x(x – 1)(x2 + 4) yields the basic equation

2x3 – 4x – 8 = A(x – 1)(x2 + 4) + Bx(x2 + 4) + (Cx + D)(x)(x – 1)

Example 3 – SolutionEquate X3 terms

2x3=Ax3+Bx3+Cx3

so 2=A+B+C (1)

Equate X2 terms

0=-AX2-Cx2+Dx2

so 0=-A-C+D (2)

Equate X terms

-4X=4AX+4BX-DX

so -4=4A+4B-D (3)

Equate X0 terms

-8=-4A (4)

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Example 3 – SolutionSolving for A

-8=-4A

A=2 (5)

Substitution of (5) into (3)

-4=4A+4B-D

-4=8+4B-D

-12=4B-D (6)

and substitution of (5) into (1)

2=A+B+C

2=2+B+C

0=B+C

B=-C (7)

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Example 3 – SolutionSubstitution of (5) into (2)

-0=-2-C+D

2=-C+D and substitution of (7) yields

2=B+D

2-B=D (8)

Now substitution of (8) and (6)

-12=4B-D

-12=4B-(2-B)

-12=4B-2+B

-12=5B-2

-10=5B

so B=-2 (9)

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Example 3 – Solution

Substitution of (9) into (7)

B=-C

So C=2 (10)

Solving (8)

2-B=D

2-(-2)=D

So D=-4

restating the problem

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Example 3 – Solution

Substituting A=2, B=-2, C=2, and D = 4, it follows that

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