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Integrating Rational Functions by Partial Fractions Objective: To make a difficult/impossible integration problem easier.

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Presentation on theme: "Integrating Rational Functions by Partial Fractions Objective: To make a difficult/impossible integration problem easier."— Presentation transcript:

1 Integrating Rational Functions by Partial Fractions Objective: To make a difficult/impossible integration problem easier.

2 Partial Fractions In algebra, you learn to combine two or more fractions into a single fraction by finding a common denominator. For example

3 Partial Fractions However, for the purposes of integration, the left side of this equation is preferable to the right side since each term is easy to integrate.

4 Partial Fractions We need a method to take and make it This method is called Partial Fractions. This method only works for proper rational fractions, meaning that the degree of the numerator is less than the degree of the denominator. This is how it works.

5 Partial Fractions Factor the denominator completely.

6 Partial Fractions Factor the denominator completely. Assign a variable as the numerator to each term of the denominator and set it equal to the original.

7 Partial Fractions Multiply by the common denominator.

8 Partial Fractions Multiply by the common denominator. Solve for A and B. To solve for A, let x = 4, which gives us To solve for B, let x = -1, which gives us

9 Example 1 Evaluate

10 Example 1 Evaluate

11 Example 1 Evaluate

12 Example 1 Evaluate

13 Example 1 Evaluate

14 Example 1 Evaluate

15 Linear Factors Linear Factor Rule. For each factor of the form, the partial fractions decomposition contains the following sum of m partial fractions where A 1, A 2, …A m are constants to be determined. In the case where m = 1, only the first term appears.

16 Example 2 Evaluate

17 Example 2 Evaluate

18 Example 2 Evaluate

19 Example 2 Evaluate

20 Example 2 Evaluate

21 Example 2 Evaluate Since there is no way to isolate A, we need to solve with a different method. Let x = 1, substitute our values for B and C and solve for A.

22 Example 2 Evaluate Since there is no way to isolate A, we need to solve with a different method. Let x = 1, substitute our values for B and C and solve for A.

23 Example 2 Evaluate

24 Example 2-WRONG Evaluate

25 Example 2-WRONG Evaluate What next? This doesn’t work!

26 Example 2-WRONG Evaluate The denominator on the right is The denominator on the left is They are not the same!!

27 Quadratic Factors Quadratic Factor Rule For each factor of the form, the partial fraction decomposition contains the following sum of m partial fractions: where A 1, A 2,…A m, B 1, B 2,…B m are constants to be determined. In the case where m = 1, only the first term appears.

28 Example 3 Evaluate

29 Example 3 Evaluate Factor by grouping

30 Example 3 Evaluate Factor by grouping

31 Example 3 Evaluate Multiply the right side of the equation and group the terms based on powers of x.

32 Example 3 Evaluate Set the coefficients from the right side of the equation equal to the ones on the left side.

33 Example 3 Evaluate Take Eq 1 – Eq 3

34 Example 3 Evaluate Take Eq 1 – Eq 3 Take new Eq + 3Eq 2

35 Example 3 Evaluate Take Eq 1 – Eq 3 Take new Eq + 3Eq 2

36 Example 3 Evaluate

37 Example 3 Evaluate

38 Example 4 Evaluate

39 Example 4 Evaluate

40 Homework Section 7.5 Page 521 1-19 odd


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