Warm up: Recall that if you require real coe cients (not complex), polynomials always factor into degree 1 or 2 parts.

Transcription

1 Today: 6.3 Partial fractions Warm up: Recall that if you require real coe cients (not comple), polynomials always factor into degree or parts. For eample, =( )( + i)( ++i) =( {z } )( + + 5) {z } allowing comple coe s using only real coe s. Factor the following polynomials (as far as possible) into factors with real coe cients. (a) +3 + (b) (c) 3 7 (d) 4 6 (e) Calculate the following (you may want to factor the denominator). + (a) d (b) d 3 (c) + d (d) + + d

2 (a) +3 +=( + )( + ) (b) 3 4 +=(3 )( ) (c) 3 7 = ( + p p 7)( 7) (d) 4 6 = ( 4)( + 4) = ( )( + )( + 4). (e) = ( ) 9( ) =( 9)( ) = ( + 3)( 3)( ) (a) R 4+4 d = R ( ) d = ( ) + C (b) R d = R + (+)(+3) d = R +3 (c) R + d = R + + d =ln + (d) R 3 ++ d = R d = ln + C. d =ln +3 + C tan ()+C Reviewing polynomial long division Calculate 3 ++ : into 3 into into ( 3 ) ( ) + + ( ) ( ) 3 + ( 3 3 ) ( )3 4 remainder So =

3 Strategies for integrating rational functions. To compute d we noted that = + ( + )( + 3) = +3, so that d =. To compute we could note that d, d =ln +3 + C. if u = +5 +6, them du = +5, so d = u du =ln u +C =ln C. Strategies for integrating rational functions 3. How about d? What if I pointed out to you that = 7( + 3) ( + )( + 3) 4( + ) ( + )( + 3) Well, then = ( + )( + 3) d = = ? d =7ln + 4ln +3 + C.

4 Strategies for integrating rational functions How could I have found that = 7 + if I didn t already know? Well, since the denominator factors as =( + )( + 3), if the fraction can be written as the sum of two fractions with linear denominators, those denominators will be ( + ) and ( + 3). So, for some A and B, weknow = A + + So we can solve for A and B as follows! B +3. Suppose A and B satisfy ( + )( + 3) = A + + B +3. Then multiplying both sides by ( + )( + 3), we get = A( + 3) + B( + ) =(A + B) +(3A +B). Comparing both sides, we know that the constant terms have to match and the coe cients on have to match (that s what it means for two polynomials to be equal!). So Plugging A =3 equation, we get 3=A + B and 3 = 3A +B. B (from the first equation) into the second 3 = 3(3 B)+B =9 3B +B =9 B. So B = 4,andso A =3 ( 4) = 7. Thus ( + )( + 3) = , as epected! +3

5 You try. Solve for A and B such that 4 + ( )( + 4) = A + B +4.. Use your previous answer to calculate R 4+ ( )(+4) d.

6 You try. Solve for A and B such that 4 + ( )( + 4) = A + B +4. Multiply both sides by ( )( + 4) to get 4 + = A( + 4) + B( ) = (A + B) +(4A B). Thus 4=A + B, sothatb =4 A, and = 4A B =4A (4 A) =5A 4. So A =3 and B =.. Use your previous answer to calculate R 4+ ( )(+4) d. 4 + ( )( + 4) d = d = 3ln +ln +4.

7 Rules for splitting rational functions If you re trying to split a rational function Q()/P (): 0. If deg(q()) deg(p ()), use long division first.. Factor the denominator P () (as far as possible) into degree and factors with real coe cients: P () =p ()p () p k ().. For each factor p i (), add a term according to the following rules. Note that p i () might appear multiple times. (a) If p i appears eactly once in the factorization: If p i is degree, add a term of the form A/p i (); If p i is degree, add aa term of the form (A + B)/p i (). Basically, add a term with p i () in the denominator and a generic polynomial of one degree less in the numerator. For eample, we would write ( + )( 3) = A + + B 3, ( + )( 3) = A + B + + C 3, and ( + )( + 3) = A + B + + C + D +3 Rules for splitting rational functions (a) If p i appears eactly once in the factorization: If p i is degree, add a term of the form A/p i (); If p i is degree, add aa term of the form (A + B)/p i (). Basically, add a term with p i () in the denominator and a generic polynomial of one degree less in the numerator. (b) If p i appears eactly n times in the factorization: If p i is degree, add terms A j /(p i ()) j for j =,,...,n; If p i is degree, add terms (A j + B j )/(p i ()) j for j =,,...,n. For eample, (split doesn t depend on numerator!) ( + )( + + ) ( + ) 4 = A F + + B + C D + E ( + + ) G ( + ) + H ( + ) 3 + I ( + ) 4.

8 You try: Split the following rational functions. Don t solve for the constants; just set it up. If need be, factor the denominator first ( + 3)( 5) ( + )( + 3) 5

9 You try: Split the following rational functions. Don t solve for the constants; just set it up. If need be, factor the denominator first ( + 3)( 5) = A +3 + B = A 3 + B = A + + B ( + )( + 3) 5 = A + B + C 3 + D + + C + D +4 + E +3 + F+ G ( + 3) + H + I ( + 3) 3 + J + K ( + 3) 4 + L + M ( + 3) 5 Definition: This split form is called partial fractions decomposition.

10 You try: Compute d. (use the previous slide to split the fraction, solve for the unknowns, and use the split form to integrate)

11 You try: Compute d. (use the previous slide to split the fraction, solve for the unknowns, and use the split form to integrate) Soln: Write so that = A 3 + B, = A( ) + B(3 ) =(A +3B) +( A B). Then =A +3B and 0= A B. So =A + 3( A )= A. Thus A = / and B =/. {z} So B d = / 3 + / d = ln 3 + ln + C.

12 You try: Compute (factor the denominator, split the fraction, solve for the unknowns, and use the split form to integrate) d.

13 You try: Compute (factor the denominator, split the fraction, solve for the unknowns, and use the split form to integrate) Soln: Write so that d. = A + B + C +, 3 +=A( + ) + (B + C) =(A + B) + C + A. Then 3=(A + B), =C, and=a. ThusB =. So d = d =ln d =ln +ln + tan ()+C We could have been a little more clever with d : Notice So = = d d (3 + ) 3 + d = d d What we got before: =ln 3 + tan ()+C. ln +ln + tan ()+C =ln ( +) tan ()+C X

14 How the integration of each part goes Suppose you ve done a partial fractions decomposition. Then each part looks like A (a + b) i or A + B (a + b + c) i (where i may be ). For calculating A (a + b) i d, let u = a + b every time. For calculating A + B (a + b + c) i d, there s a little more to be done. Whenever i =, you will split this into two fractions: one which has some constant times d d (a + b + c) =a + b in the numerator (so u = a + b + c), and one which has a constant in the numerator (so I can use d d tan ( ) =/( + ), possibly after completing the square). Integrating (A + B)/(a + b + c) + Eample: + d. As I said, I want to break this up into two fractions, one which has some constant times d d ( + ) = in the numerator (so u = +), and one which has a constant in the numerator (so I can use d d tan () =/( + ): so that + + = + + d = + + +, + d + = ln + + tan ()+C. + d

15 Integrating (A + B)/(a + b + c) 3 +5 Eample: +4 d. Again, I want one u = +4(so du =) and one tan (u) integral = So = d = d (/) +. (/) + d. For the first, let u = +4,sodu =, andthus 3 +4 d = 3 u du = 3 ln +4 + C; for the second, let u = /, sodu = d, andthus 5 4 (/) + d = 5 u + du = 5 tan (/) + C. Integrating (A + B)/(a + b + c) Eample: d. Complete the square of the denominator to get this fraction into the form of the previous two eamples: +4 +7=( + ) 4+5=( + ) +. Now let u = +(so that du = d and = u ): d = ( + ) + d = u u + du. Now proceed as before! u u + du = u u + du u + du = ln u + tan (u)+c = ln (+) + tan (+)+C.

16 Integrating (A + B)/(a + b + c) i If there s nothing obvious to be done, even for i>, complete the square of the denominator. In general, this goes as follows:. Factor out a i : rewrite the denominator as a i ( + + ) i, where = b/a and = c/a.. Complete the square of the rest: + + =( + /) +( /4) =( + /) + where = p /4. (real since the denom. s not factorable) 3. Make the constant term by factoring out :! + / i (a +b+c) i = a i ((+ /) + ) i = a i i Let u = + /. Use = / to rewrite the numerator. (Don t try to memorize these equations. Just know that the process works every time.) You try: Compute d. [Hint:. The degree of the numerator is not less than the degree of the denominator. So reduce first. You should end up with B+D something that lookes like A +. Since the denom. is not factorable, get it into the form k((f()) + ) as on the previous slide. Then let u = f(). ]