Partial Fractions

We continue the discussion we started last lecture about integrating rational functions. We definedarationalfunctionastheratiooftwopolynomials:
P(x)
Q(x)
Welookedattheexample
13
x ? 1+ x +2 dx = ln |x ? 1| + 3ln |x +2| + c
Thatsameproblemcanbedisguised:
1+3 =(x + 2) + 3(x ? 1)= 4x ? 1
x ? 1 x +2 (x ? 1)(x + 2) x2 + x ? 2
whichleavesustointegratethis: ?
4x ? 1 dx = ???
x2 + x ? 2 P(x)
Goal: we want to figure out a systematic way to split into simpler pieces.
Q(x)
First,wefactorthedenominator Q(x).
4x ? 1= 4x ? 1= A + B
x2 + x ? 2(x ? 1)(x + 2) x ? 1 x +2
There’saslowwaytofind A and B. You can clear the denominator by multiplying through by (x ? 1)(x + 2): (4x ? 1) = A(x + 2) + B(x ? 1)
Fromthis,youfind 4= A + B and ? 1=2A ? B
Youcanthensolvethesesimultaneouslinearequationsfor A and B. This approach can take a very longtimeifyou’reworkingwith3,4,ormorevariables.
There’s a faster way, which we call the “cover-up method”. Multiply both sides by (x ? 1):
4x ? 1 B x +2 = A + x + 2(x ? 1)
Set x =1 tomakethe B termdropout:
4 ? 1= A
1+2
A =1
1
Lecture 29 18.01 Fall 2006
The fastest way is to do this in your head or physically cover up the struck-through terms. For instance,toevaluate B: 4x ? 1 A? B (x ? 1)??? ? x ? 1(x + 2)
(x +2) =??+??? ? Implicitly,wearemultiplyingby (x + 2) andsetting x = ?2. This gives us
4(?2) ? 1= B = B =3
?2 ? 1 ?
What we’ve described so far works when Q(x) factors completely into distinct factors and the degreeof P islessthanthedegreeof Q.
Ifthefactorsof Q repeat, we use a slightly different approach. For example:
x2 +2 ABC
=+ +
(x ? 1)2(x + 2) x ? 1(x ? 1)2 x +2 Usethecover-upmethodonthehighestdegreetermin (x ? 1).
x2 +1 12 +2
= B +[stuff](x ? 1)2 == B = B =1
x +2 ? 1+2 ? Implicitly,wemultipliedby (x ? 1)2,thentookthelimitas x ? 1. C can also be evaluated by the cover-up method. Set x = ?2 toget
x2 +2 2 = C +[stuff](x +2) = (?2)2 +2 = C = C =2
(x ? 1) ? (?2 ? 1)2 ? 3
Thisyields
x2 +2 A 12/3
=+ +
(x ? 1)2(x + 2) x ? 1(x ? 1)2 x +2 Cover-upcan’tbeusedtoevaluateA.Instead,pluginaneasyvalueofx: x =0. 2 A 11 1
(?1)2(2) = ?1+1+ 3 =? 1=1+ 3 ? A =? A =3
Nowwehaveacompleteanswer:
x2 +2 112
= ++
(x ? 1)2(x + 2) 3(x ? 1) (x ? 1)2 3(x + 2)