74 The Method of Partial Fractios I algebra oe speds much time fidig commo deomiators ad thus simplifyig ratioal epressios For eample: + + + 6 5 + = + = = + + + + + ( )( ) 5 It may the seem odd to be watig to do this backwards, startig with ( )( + ) ad tryig to fid the two simpler ratioal epressios whose sum is that epressio, + Seeig why oe + 5 might wat to do this becomes obvious if you cosider the itegral d I that form it s ( )( + ) difficult to see how to approach solvig this itegral But rewritte as the sum of those simpler epressios, we ca solve the itegral easily: 5 d d d = + d = l + l + + C + + + 5 ( )( + ) d = l + l + + C d I the above ad i what follows, we ll make use of a old result: l a C + a + du To see why, just make the substitutio u a, du = d, to get l u C l a C u + We ll et study the techique of partial fractio decompositio, whereby ratioal epressio are decomposed ito a sum of simpler epressios Partial Fractio Decompositio Step : The very first step is to make sure that the degree of the umerator is less tha the degree of the deomiator If ot, use polyomial log divisio The result will be a polyomial plus a ratioal epressio with umerator of lesser degree tha its deomiator Later we ll see eamples where we apply that first step But for ow, we ll assume that s bee doe The et step, as the above eample suggests, is to factor the deomiator completely
Partial Fractio Decompositio Step : Factor the deomiator completely The above two steps apply to all partial fractio problems Now we ll start eamiig special cases Whe the deomiator factors ito distict liear factors, the decompositio takes the form f( ) b b b + + a a a a a a ( ), for some costats b i The above is ot just a equatio, but a idetity It s true for all for which it s defied The problem is to determie all the costats b, b,, b I priciple this is possible to do, sice each time you substitute oe particular value of you get oe equatio ivolvig the b i Doig this times should result i liear equatios i ukows This souds like a bit of a hassle though, sice liear systems are so tedious to solve by had However there s a ice short-cut you ca apply wheever you have all distict liear factors i the deomiator We ll illustrate this short-cut i the followig eample + 5 Eample Fid the partial fractio decompositio of + 5+ 6 The deomiator factors ito distict liear factors: So we seek a partial fractio decompositio of the form: + 5 + 5 = + 5 + 6 + + + 5 a b + + + + Multiply both sides by the deomiator to cacel all the deomiators: + 5= a + + b + The above is a idetity true for all So let s pick particularly ice values of to plug i: ( ) ( ) ( ) ( ) + 5 = a + + b + = + 5 = a + + b + = a+ 0 a = = + 5 = a + + b + = 0 b b= We picked values of that made all but oe term vaish, so it was quick ad easy to solve! Thus: + 5 = + 5 + 6 + +
I our et eample, we ll use the above result, but we ll start with a itegral + 5 Eample d =? Usig the partial fractio decompositio from Eample, + 5+ 6 + 5 d d d = d = = l + l + + C + 5 + 6 + + + + I our et eample, there are more liear factors; yet the problem is ot much more difficult tha our previous decompositio problem 4 0 Eample Fid the partial fractio decompositio of 4 The deomiator factors ito distict liear factors: 4 0 4 0 4 0 = = 4 + ( 4) Ad so, the decompositio we seek is of the form: 4 0 a b c + + + Clear the deomiators, ad begi substitutig i ice values of : 4 0= a + + b + c + = 0 4 0 = a 0 + 0 + b 0 + c 0 4 = 4 a a = = 4 + 0 = a 0+ b + c 0 4 = 8 b b= = 4 0 = a 0 + b 0 + c 4 6 = 8 c c= 4 0 Thus, = + 4 + 4 0 Eample 4? d = Usig the partial fractio decompositio from Eample, 4 4 0 d d d + + d = d 4 + = + = l + l + l + C
The et case we ll cosider is the case where the deomiator has a repeated liear factor: f( ) b b b = + + + ( a) a ( a) ( a), for some costats b i Eample 5 Fid the partial fractio decompositio of + + ( + ) We seek a decompositio of the form: Clearig the deomiators, we get: + + + a b c ( + ) + ( + ) ( +) + + + + + a b c = 4 + = a 0 + b 0 + c c= ( ) ( ) () () = 0 0 + 0 + = a 0 + + b 0 + + = 4a+ b+ = a+ b = + = a + b + = a+ b+ = a+ b You ll otice that we foud c as usual, but the we were forced to choose less ice values of, which did t etirely elimiate the other variables; but at least resulted i a simple liear system i two variables (there are ifiitely may values of oe ca choose to substitute; so pick the oes that result i the simplest equatios) Notice that we immediately substituted the c = value we foud right back ito the other equatios With our two equatio liear system, simply subtract the secod equatio from the first to obtai a = The substitute this value ito the last equatio to fid b = Thus : + + = + + + + + ( ) Eample 6 + + d =? Usig the results of Eample 5, our itegral becomes: ( + ) d = + d = d + + + d ( + ) + ( + ) ( + ) + ( + ) ( + ) Those last two itegrals ca be solved with the substitutio u, du = d: du du du u u = l + + u du u du l C u = u + + + u + + = l + + ( + ) ( + ) + C d= l + + + C d ( + ) + ( + )
Combiatios of the above two cases occur For eample, cosider the form of the followig partial fractio decompositio: + + 4 0 6 a b c d + + ( + )( )( ) + ( ) Notice how it aturally combies the two cases of partial fractio decompositio we ve see so far, the case of distict liear factors ad the case of a repeated liear factor Eample 7 Fid the partial fractio decompositio of + + 4 0 ( + )( )( ) 6 4 + 0 + 6 a b c d + + This decompositio is of the form ( + )( )( ) + ( ) Clearig the deomiators, we obtai: ( ) 4 0 6 + + = a + b + + c + + d + ( ) = + 80 6 + 6 = a 0 + b 4 + c 0 + d 0 = 4 b b = = + 80 + 6 + 6 = a 4 5 + b 0 + c 0 + d 0 00 = 00 a a = = 08 + ()( ) 80 9 + 6 = a 0 + b 0 + c 0 + d 5 5 = 5 d d = We ve foud all the costats ecept c Settig = 0 ad substitutig i the costats we ve already foud, we get: 0+ 0 0+ 6 = a ( )( ) + b( )( ) + c( )( )( ) + d ( )( ) 6= ( )( )( 9) + ( )( 9) + c( ) + ( )( ) 6 = 6 + 54 + c 4 60 = c c = 5 Ad so, + + + 4 0 6 5 ( + )( )( ) + ( ) Eample 8 + + 4 0 6 d =? Usig the results of the last eample: ( + )( )( ) + + d = + + 4 0 6 5 ( + )( )( ) + ( ) d
5 d d d + + d = + 5 + ( ) d ( ) + + 4 0 6 + + ( + )( ) ( ) ( ) = l + + l 5 l + +C d = l + + l 5l + C Without eterig ito the realm of the comple umbers, oe sometimes ecouters irreducible quadratic polyomials The partial fractio decompositio form to use for irreducible quadratic factors i the deomiator is easiest to show i the most geeral case, whe that irreducible quadratic factor is repeated: f ( ) m + b m + b m + b + + a b c a + b + c + + a + b + c a + b + c ( ) Here i the deomiator we are give a repeated irreducible quadratic factor, a + b + c, of multiplicity We are to solve for all the costats m, m,, m ad b, b,, b 5 + + Eample 9 Fid the partial fractio decompositio of ( )( + ) This deomiator has a liear factor ad a irreducible quadratic factor: Clearig deomiators: 5 + + a m+b + ( )( + ) ( ) 5 + + = a + + m+ b = 5 + + = a + + m+ b 0 0 = a a = 5 () ( ) ( m+ )( ) = 0 0 + 0 + = 5 + 0 + b 0 = 5 b b= = 5 + = 5 + + 6 = 0 + m 4 m= 0 Ad so, 5 + + 5 + ( )( + ) Eample 0 5 + + d =? Usig the results of the last eample: ( )( + )
d d + + 5 + + 5 d d = 5 d + = 5l + ta + C ( )( + ) For our et eample, we eed aother solutio techique If we try as before, we will ed up with a full blow system of 4 liear equatios with 4 ukows to solve That certaily ca be doe, but it s a bit too tedious to do by had There is yet aother techique that we could have used to solve all our previous eamples However the approach we took above was less hassle But ot this time! Eample Fid the partial fractio decompositio of + + + ( + ) This deomiator cosists of a irreducible quadratic factor of multiplicity : Clearig deomiators, as we ve doe before: + + + m+ b + c ( + ) + ( + ) m b + + + + + +c This time, istead of pluggig i values for, multiply out the right side of the equatio, ad write it as a cubic polyomial with variable i descedig order: + + + = m + m + b + b + + c + + + = m + b + m+ + b+c ( ) The fact that this is a idetity meas that the polyomial o the left eactly equals the polyomial o the right, which meas that all the correspodig coefficiets are equal! Thus, m=, b=, m+ =, b+ c = m=, b=, + =, + c= m=, b=, = 0, c= Ad so, + + + + ( + ) + ( + ) Eample + + + d =? Usig the results of the last eample: ( + ) + + d d d d + d d ( ) ( ) + + + + + + ( + ) For the first itegral let u = +, du = d, ad for the last let ta, d sec d = θ = θ θ, to get:
du sec θ dθ sec θ dθ = + + = + + ta l u ta u ( ta θ + ) ( sec θ) dθ = l( + ) + ta + l ( ) ta cos θdθ = + + + sec θ + cosθ = l + + ta + θ = l + + ta + + cos θ ( ) d ( ) ( ) dθ = l ( + ) + ta + d siθ θ + cos θ dθ = l ( + ) + ta + θ + + C siθcosθ = l ( + ) + ta + ta + + C = l ( + ) + ta + siθ cosθ + C = l ( + ) + ta + + = l ( + ) + ta + + + + + C C That problem was a good itegratio review! There is yet aother way to do our last partial fractio decompositio, for those ot afraid of imagiary umbers It s a sort of hybrid techique, combiig aspects of both of our previous approaches Notice how we make use of the fact that two comple umbers are equal if ad oly if both their real parts are equal ad their imagiary parts are equal Eample (Eample, revisited) Fid the partial fractio decompositio of This deomiator cosists of a irreducible quadratic factor of multiplicity : + + + + + m b c ( + ) + ( + ) + + + ( + ) Clearig deomiators, as we ve doe before; but this time, lets let = i: + + + = m+ b + + + c + + = i i + i+ = mi+ b + + i+ c + 0i = c + i c= & = 0 = 0 = 0 + b 0 + 0i+ = b+ b= = + = m+ + + 0 + 6 = m+ + m= As before, we ve obtaied the decompositio + + + + ( + ) + ( + ) I our et eample, partial fractios do t arise util after a substitutio
Eample 4 + e e + e 5e 8e 4 ed=? Let u = e, du = e d, ad our itegral becomes: 5 8 4 5 8 4 5 8 4 u u+ u u+ u u+ = du = du = du u u + u u( u u+ ) u( u )( u ) Now we ca do a partial fractio decompositio of the itegrad: Clearig the deomiators: 5u 8u+ 4 a b c + u u u u u u ( ) () = 5 8 4 u u+ = a u u + bu u + cu u u = 0 0 0 + 4 = a + b 0 + c 0 4 = a a = u = 5 8 + 4 = a 0 + b + c 0 = b b= u = 0 6 + 4 = a 0 + b 0 + c 8 c c= 4 Thus, 5u 8u+ 4 4 = +, ad our itegral becomes: u u u u u u 5u 8u + 4 4 du du du = du = + du = du + 4 du u u u u u u u u u 5e 8e 4 ( ) = l u lu + 4l u + C = le le + 4le + C + e e + e ed= l e + 4l e + C Eample 5 5cos si 8cos si + 4cos si si + si d =? + = cos d 5si 8si 4 si si + si Lettig u = si, du = cos d, our itegral becomes 5u 8u + 4 5u 8u 4 du + u u + u u u u = = du This is the same itegrad as i Eample 4, so we ll write dow the solutio we previously obtaied: = l u l u + 4 l u + C 5cos si 8cos si + 4cos si si + si d = lsi lsi + 4lsi + C
Whe the degree of the polyomial i the umerator is greater tha or equal to that of the deomiator, we must first perform log divisio, before fidig the partial fractio decompositio Eample 6 + 9 + 7 6 + 7+ 0 d =? Sice the degree of the umerator eceeds that of the deomiator, we begi with log divisio: + 7 + 0 + 9 + 7 6 ( 0) + + 6 ( 4 0) + 4 + 9 + 7 6 4 = + + + + + + 7 0 7 0 Net, we perform a partial fractio decompositio for the remaiig ratioal epressio: Clearig deomiators: + 4 + 4 a b = + 7 + 0 + + 5 + + 5 ( ) ( ) + 4 = a + 5 + b + = + 4 = a + 5 + b 0 = a a = 4 = 5 5 + 4 = a 0 + b 5 + 9 = b b= Ad so we have that + 9 + 7 6 4 = + + + + + 7 0 5 Thus our itegral becomes: + + d d + 7 + 0 + + 5 + + 5 9 7 6 4 d = d d d 4 + = + = + 4 l + l + 5 + C + + + 7+ 0 9 7 6 d = + 4l + l + 5 + C I summary, to apply the method of partial fractios, first make sure the degree of the umerator is less tha or equal to the degree of the deomiator If this is t the case, perform log divisio The remaider (if ay) ca the be decomposed ito partial fractios Net, factor the deomiator
completely Each factor of the deomiator gives rise to partial fractio(s), as follows I the followig chart, : Factor i the deomiator Resultig partial fractio decompositio ( a) b b b + + + a a ( ) ( a) ( a + b + c) (irreducible) m + b m + b m + b + + + a + b + c a + b + c a + b + c