ENGI 44 Engineering Mthemtics Five Tutoril Exmples o Prtil Frctions 1. Express x in prtil rctions: x 4 x 4 x 4 b x x x x Both denomintors re liner non-repeted ctors. The cover-up rule my be used: 4 4 4 4 1; b 1 4 4 Thereore 4 1 1 x 4 x x x Among other uses, it then ollows tht 4 x dx ln x ln x C ln C x 4 x x ln x ln A ln A x x
ENGI 44 Prtil Frctions Exmples Pge o 6. Express x in prtil rctions: x x 1 x x x x 1 x 1 b c x x x x 1 x 1 x x 1 x 1 1 All three denomintors re liner non-repeted ctors. The cover-up rule my be used: b c 0 1 1 0 0 10 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 Thereore 1 1 x 1 1 1 x 1 1 1 1 x x 1 x x 1 x 1 While it is true tht one cn conclude rom this prtil rction expression tht x 1 dx ln x ln x 1 ln x 1 C ln xx 1 x 1 C ln x x C x x this integrl cn be ound more quickly by using u x dx ln u x C u x u x or ll dierentible unctions with ux x x
ENGI 44 Prtil Frctions Exmples Pge o 6. Express x in prtil rctions: x 1 x x x 1 1 bx c x x x x 1 x x 1 Note tht the polynomil in the numertor o prtil rction must be o order one less thn tht o the denomintor o tht prtil rction. Liner denomintors need constnt numertors, while qudrtic denomintors require liner numertors. OR Only the irst denomintor is liner non-repeted ctor. used to ind : 1 1 1 0 0 1 1 One o the stndrd methods must be used to ind b nd c. Clering the denomintors (nd replcing by +1 ): 1 1 x 1 bx c x 1 b x cx 1 or ll x Mtching coeicients o Mtching coeicients o [The coeicients o x : 0 1b b 1 x 1 : 0 c c 0 The cover-up rule my be 0 x lredy mtch, becuse we ound = 1 by the cover-up rule.] Setting x 1: 1 1 b c 1 A Setting x 1: 1 1 b c 1 B A B c 0 c 0 in A or B b 1 [Note tht setting x = 0 provides no new inormtion, becuse we ound = 1 by the cover-up rule.] Thereore 1 1 x x x x x x 1 Among other uses, it then ollows tht 1 1 x dx ln x ln x 1 C ln C x x x 1
ENGI 44 Prtil Frctions Exmples Pge 4 o 6 4. Express Fs in prtil rctions: F s 6s s8 s 1 s 1 s 6s s 8 b c s 1 s 1 s s 1 s 1 s All three denomintors re liner ctors tht re not repeted. The cover-up rule my be used in ll three cses. c 111 6 1 1 8 1 1 0 6 5 61 1 8 4 1 1 1 1 1 6 8 1 1 18 Thereore F s 17 6 6s s8 5 17 6 s 1 s 1 s s 1 s 1 s I one does not wish to employ the cover-up rule, then strt by clering the denomintors: 6s s 8 s 1 s b s 1 s c s 1 s 1 Setting s 1 yields 6 8 6 0 0 Setting s 1 yields 6 8 0 b 0 Setting s yields 4 48 0 0 c The sme vlues or, b nd c then ollow.
ENGI 44 Prtil Frctions Exmples Pge 5 o 6 5. Express Fs in prtil rctions: F s 5s 7s s 1 s1 s 1 All lower positive integer powers o repeted ctor must pper. In this cse, there must be seprte terms or s 1 nd s 1. The reson or this rule is on the next pge. Where the [non-repeted] denomintor o prtil rction is n n th order polynomil, the corresponding numertor must be n (n 1) th order polynomil. The simple cover-up rule cnnot be used in either cse, (but modiied version cn be used to ind b). 5s 7s s 1 b cs d s 1 s 1 s1 s 1 s 1 5s 7s s 1 s 1 s 1 b s 1 cs d s 1 Setting s 1 yields 5 7 1 0 b 0 b Setting s 0 (or, equivlently, mtching coeicients o 0 0 01 d d At this point we hve severl choices: Mtch two o the remining three coeicients, or Substitute two more vlues o s, or Some combintion o the bove two methods. In this exmple I shll choose to mtch coeicients o s : 5 0 c c 5 0 s ) yields s nd s : 7 d c 10 c 5 nd d 4 5s 7s s 1 s 4 1 s 1 s 1 s 1 s s 1 s.
ENGI 44 Prtil Frctions Exmples Pge 6 o 6 Explntion o Repeted Prtil Frctions Where the denomintor o prtil rction is n n th order polynomil, the corresponding numertor must be n (n 1) th order polynomil. In question 5 prtil rction is repeted nd so should be o the orm: ms n ms m m n ms 1 m n ms1 n m s 1 s 1 s 1 s 1 s 1 ms n b s1 s 1 s1, where m nd b n m Multiply repeted liner ctors re treted in similr wy. For exmple, mx nx px q b c d 4 4 x k x k x k x k x k Repeted qudrtic ctors re lso treted in similr wy: mx nx px q x b cx d x rx s x rx s x rx s Why does the simple cover-up rule work (or non-repeted liner ctors)? s bs Consider this more generl cse: s k g s s k g s where nd k re constnts, gs is n (n) th order polynomil in s tht does not hve s k gk is not zero), s is polynomil in s o order t most (n) nd bs s ctor, (so tht is polynomil in s o order n 1 contining n coeicients to be determined. s g s bss k s Let s = k then k g k b k k k g k 0 which is just the originl rction evluted t s k gk k with the zero ctor s k covered up. Return to your previous pge Creted 006 0 15 nd most recently modiied 017 07 17 by Dr. G.H. George