1. Partial Fractions

2. 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

3. Partial Fractions

4. 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

5. 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, x 2 – 5x + 6, and to find the partial fractions Partial fraction decomposition

6. Partial Fractions

7. Linear Factors

8. Example 1 – Distinct Linear Factors Write the partial fraction decomposition for Solution: Because x 2 – 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.

9. Example 1 – Solution To solve for A, let x = 3 and obtain 0x+1 = Ax+Bx-2A – 3B Equating x 0 terms 1 = -2A -3B(1) and Equating x 1 terms 0 = A+B so-A=B(2) Substitution of (2) into (1) yields 1=2B-3B 1=-1B thus-1=B(3) andA=1(4) cont’d

10. 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 cont’d

11. Quadratic Factors

12. Example 3 – Distinct Linear and Quadratic Factors Find Solution: Because (x 2 – x)(x 2 + 4) = x(x – 1)(x 2 + 4) you should include one partial fraction for each factor and write Multiplying by the least common denominator x(x – 1)(x 2 + 4) yields the basic equation 2x 3 – 4x – 8 = A(x – 1)(x 2 + 4) + Bx(x 2 + 4) + (Cx + D)(x)(x – 1)

13. Example 3 – Solution Equate X 3 terms 2x 3 =Ax 3 +Bx 3 +Cx 3 so 2=A+B+C (1) Equate X 2 terms 0=-AX 2 -Cx 2 +Dx 2 so 0=-A-C+D(2) Equate X terms -4X=4AX+4BX-DX so -4=4A+4B-D(3) Equate X 0 terms -8=-4A(4) cont’d

14. Example 3 – Solution Solving 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) cont’d

15. Example 3 – Solution Substitution 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) cont’d

16. 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 cont’d

17. Example 3 – Solution Substituting A=2, B=-2, C=2, and D = 4, it follows that cont’d