MAT01B1: Integration of Rational Functions by Partial Fractions

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1 MAT01B1: Integration of Rational Functions by Partial Fractions Dr Craig 1 August 2018

2 My details: Dr Andrew Craig Consulting hours: Monday 14h40 15h25 Thursday 11h20 12h55 Friday 11h20 12h55 Office C-Ring or google Andrew Craig maths

3 Tut 3 Prescribed questions have been posted on Blackboard. Saturday class this week D1 LAB 108, 09h00 to 12h00 Working on all integration techniques that we have learnt so far.

4 Consider the following operation involving two rational functions: 3 x 7 1 x + 2 = 3(x + 2) 1(x 7) (x 7)(x + 2) 2x + 13 = x 2 5x 14 It is simple to go from the original subtraction to the final answer. Being able to go in the opposite direction can be very useful for simplifying integrals.

5 Rational functions Let P (x) be a polynomial. That is, P (x) = a n x n + a n 1 x n a 1 x + a 0 with a n 0. Then we say that the degree of P is n and we can write deg(p ) = n. A rational function f = P (x) is called Q(x) proper if the degree of P is less than the degree of Q.

6 Improper rational functions A rational function f(x) = P (x) Q(x) is improper if deg(p ) deg(q). In this case we can perform long division to get the following: f(x) = P (x) Q(x) = S(x) + R(x) Q(x)

7 Example: 2x 3 + 7x 2 + 2x + 9 x =? 2x 3 + 7x 2 + 2x + 9 x = 2x x 12 x 2 + 3

8 Example: Use long division to obtain a proper rational function from x 3 + x x 1 Hence, find x 3 + x x 1 dx

9 Solution: x 3 + x x 1 = x2 + x x 1 Therefore x 3 + x x 1 dx = = x3 3 + x2 2 ( x 2 + x ) x 1 + 2x + 2 ln x 1 + C dx

10 Partial fraction decomposition of proper rational functions

11 Simplifying rational integrands The next four cases describe how to simplify an integrand that is a rational function P (x) Q(x). The four cases that follow each deal with different forms of the denominator Q(x). Note that each case requires the rational function to be proper. If the integrand is an improper rational function, long division must first be performed and then one of the following four cases can be applied.

12 Case I: Q(x) is a product of distinct linear factors In this case we can write Q(x) = (a 1 x + b 1 )(a 2 x + b 2 )... (a k x + b k ) where no factor is repeated (and no factor is a constant multiple of any other). We have: P (x) Q(x) = A 1 a 1 x + b A k a k x + b k

13 Consider the integral 2x + 13 x 2 5x 14 dx We could try to integrate using u-substitution, but this won t quite work. So, we observe that 2x + 13 x 2 5x 14 = 2x + 13 (x 7)(x + 2) = A x 7 + B x + 2 Then we solve for A and B and split up the original integral.

14 Example: Evaluate = [ x x 2 + 2x 1 2x 3 + 3x 2 2x dx 1 2x ] 1 x + 2 dx = 1 2 ln x ln 2x 1 1 ln x C 10

15 Example: Show that dx x 2 a = 1 2 2a ln x a x + a + C

16 Case II: Q(x) is a product of linear factors, some of which are repeated If there is a repeated linear factor, e.g. then we get P (x) Q(x) = A 1 a 1 x + b 1 + (a 1 x + b 1 ) r A 2 (a 1 x + b 1 ) A r (a 1 x + b 1 ) r

17 Example of Case II Find x 4 2x 2 + 4x + 1 x 3 x 2 x + 1 dx (Hint: first perform long division and then factorise the denominator.) Solution: x x 2 x 1 + ln x 1 ln x C

18 A useful homework exercise: Use u-substitution to prove that: dx = 1 ( x ) x 2 + a 2 a arctan + C a (Hint: first factor out a 2.)

19 Case III: Q(x) contains irreducible factors, none of which is repeated Remember: an irreducible quadratic factor is one for which b 2 4ac < 0. If Q(x) has ax 2 + bx + c where b 2 4ac < 0 then we get Ax + B ax 2 + bx + c

20 Example: Give the partial fraction decomposition of the following rational function: x (x 2)(x 2 + 1)(x 2 + 4)

21 Example of Case III: I = 2x 2 x + 4 x 3 + 4x dx First we rewrite the integrand as: 2x 2 x + 4 x 3 + 4x = 2x2 x + 4 x(x 2 + 4) = A x +Bx + C x After solving for A, B and C we get: 1 I = x dx + x x dx 1 x dx.

22 Common error Note that: 2x 2 + x (x 2 1)(x + 2) Ax + B x because x 2 1 is not an irreducible quadratic factor of the denominator. C x + 2 The correct partial fraction decomposition is 2x 2 + x (x 2 1)(x + 2) = A x 1 + B x C x + 2.

23 Example 6 (pg 490): Evaluate First step: ( 1 + 4x 2 3x + 2 4x 2 4x + 3 dx ) x 1 4x 2 4x + 3 dx Then, 4x 2 4x + 3 = (2x 1) and let u = 2x 1.

24 Case IV: Q(x) contains a repeated irreducible quadratic factor (ax 2 + bx + c) r is in Q(x) with b 2 4ac < 0. Then we get r fractions with linear factors in the numerator. A 1 x + B 1 (ax 2 + bx + c) A r x + B r (ax 2 + bx + c) r

25 Example: p.f. decomposition Write out the partial fraction decomposition of x 3 + x x(x 1)(x 2 + x + 1)(x 2 + 1) 3

26 Example (Case IV) Evaluate: 1 x + 2x 2 x 3 I = x(x 2 + 1) 2 dx 1 x + 2x 2 x 3 x(x 2 + 1) 2 = A x +Bx + C x Dx + E (x 2 + 1) 2

27 Solution: I = ( 1 x x + 1 x ) x (x 2 + 1) 2 dx = ( 1 x x x x ) x dx (x 2 + 1) 2 = ln x ln(x2 +1) arctan x 2(x 2 + 1) +C

28 Summary of the four cases Case I: Q(x) is a product of distinct linear factors Case II: Q(x) is a product of linear factors, some of which are repeated Case III: Q(x) contains irreducible factors, none of which is repeated Case IV: Q(x) contains a repeated irreducible quadratic factor Since every polynomial Q(x) can be written as a product of (possibly repeated) linear and irreducible quadratic factors, these four cases cover all possible denominators Q(x).

29 Next time: In the lecture on Tuesday we will see how to use a substitution to rationalize an expression and thereby turn the expression into one to which we can apply the method of partial fractions. (You will need this method for the last few prescribed exercises.)

Math123 Lecture 1. Dr. Robert C. Busby. Lecturer: Office: Korman 266 Phone :

Math123 Lecture 1. Dr. Robert C. Busby. Lecturer: Office: Korman 266 Phone : Lecturer: Math1 Lecture 1 Dr. Robert C. Busby Office: Korman 66 Phone : 15-895-1957 Email: rbusby@mcs.drexel.edu Course Web Site: http://www.mcs.drexel.edu/classes/calculus/math1_spring0/ (Links are case