Partial FractionsLesson 7.3, page 737

Objective: To decompose rational expressions into partial fractions.

Review: Rational Expressions

rational function – a quotient of two polynomials

where p(x) and q(x) are polynomials and where q(x) is not the zero polynomial.

( )( )

( )

p xf x

q x

ReviewAddingSubtracting Rationals

2

4 18

3 3 9

x

x x x

Definition

2

5 6 3 2

( 3)( 2) 3 6

y y

y y y y y

In the problem above, the fractions on the right arecalled “partial fractions”. In calculus, it is often helpfulto write a rational expression as the sum of two ormore simpler rational expressions.

What is decomposition of partial fractions?

Writing a more complex fraction as the sum or difference of simpler fractions.

Examples:

Why would you ever want to do this? It’s EXTREMELY helpful in calculus!

22

35353412

43

xxx

x

baba

Example 1

Decompose into partial fractions.

Begin by factoring the denominator:(x + 2)(2x 3). We know that there are constants A and B such that

To determine A and B, we add the expressions on the right…giving us…

2

4 13

2 6

x

x x

4 13

( 2)(2 3) 2 2 3

x A B

x x x x

Equate the numerators: 4x 13 = A(2x 3) + B(x + 2)

Since the above equation containing A and B is true for all x, we can substitute any value of x and still have a true equation.

If we choose x = 3/2, then 2x 3 = 0 and A will be eliminated when we make the substitution. This gives us

4(3/2) 13 = A[2(3/2) 3] + B(3/2 + 2) 7 = 0 + (7/2)B.

B = 2.

4 13 (2 3) ( 2)

( 2)(2 3) ( 2)(2 3)

x A x B x

x x x x

If we choose x = 2, then x + 2 = 0 and B will be eliminated when we make the substitution. So, 4(2) 13 = A[2(2) 3] + B(2 + 2)

21 = 7A.

A = 3. The decomposition is as follows:

2

4 13 3 2 3 2 or

2 6 2 2 3 2 2 3

x

x x x x x x

Example 1 continued

Check Point 1, page 740Find the partial fraction decomposition.

5 1

( 3)( 4)

x

x x

What if one on the denominators is a linear term squared?

This is accounted for by having the nonsquared term as one denominator and having the squared term as another denominator.

What if one denominator is a linear term cubed? There would be 3 denominators in the decomposition:

32 )()()( term

C

term

B

term

A

Example 2: Decompose into partial fractions.

2

2

7 29 24

(2 1)( 2)

x x

x x

2

2 2

7 29 24

(2 1)( 2) 2 1 2 ( 2)

x x A B C

x x x x x

Example 2 continuedNext, we add the expression on the right:

Then, we equate the numerators. This gives us

Since the equation containing A, B, and C is true for all of x, we

can substitute any value of x and still have a true equation. In order to have 2x – 1 = 0, we let x = ½ . This gives us

2 2

2 2

7 29 24 ( 2) (2 1)( 2) (2 1)

(2 1)( 2) (2 1)( 2)

x x A x B x x C x

x x x x

2 27 29 24 ( 2) (2 1)( 2) (2 1)x x A x B x x C x

2 21 1 17( ) 29 24 ( 2) 0.

2 2 2A

Solving, we obtain A = 5.

Example 2 continued

In order to have x 2 = 0, we let x = 2.

Substituting gives us

To find B, we choose any value for x except ½ or 2

and replace A with 5 and C with 2 . We let x = 1:

27(2) 29(2) 24 0 (2 2 1)

2

C

C

2 27 1 29 1 24 5(1 2) (2 1 1)(1 2) ( 2)(2 1 1)

2 5 2

1

B

B

B

Example 2 continued

The decomposition is as follows:

2

2 2

7 29 24 5 1 2.

(2 1)( 2) 2 1 2 ( 2)

x x

x x x x x

Check Point 2, page 741Find the partial fraction decomposition.

2

2

( 1)

x

x x

Check Point 3, page 743Find the partial fraction decomposition.

2

2

8 12 20

( 3)( 2)

x x

x x x