082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

1.7 Partial fractions 39
1.7 PARTIALFRACTIONS
Givenasetoffractions,wecanaddthemtogethertoformasinglefraction.Forexample,
inExample 1.32 wesaw
2
x +1 + 4
x +2 = 2(x+2)+4(x+1)
(x +1)(x +2)
= 6x+8
x 2 +3x+2
Alternatively,ifwearegivenasinglefraction,wecanbreakitdownintothesumofeasier
fractions. These simple fractions, which when added together form the given fraction,
6x+8
arecalledpartialfractions.Thepartialfractionsof
x 2 +3x+2 are 2
x +1 and 4
x +2 .
Whenexpressingagivenfractionasasumofpartialfractionsitisimportanttoclassifythefractionasproperorimproper.Thedenominatoristhenfactorizedintoaproduct
of factors which can be linear and/or quadratic. Linear factors are those of the form
x
ax + b, for example 2x − 1, + 6. Repeated linear factors are those of the form
2
(ax +b) 2 , (ax +b) 3 andsoon,forexample (3x −2) 2 and (2x +1) 3 arerepeatedlinear
factors.Quadraticfactorsarethoseoftheformax 2 +bx+c,forexample2x 2 −6x+1.
1.7.1 Linearfactors
We can calculate the partial fractions of proper fractions whose denominator can be
factorized into linear factors.The following steps areused:
(1) Factorize the denominator.
(2) Eachfactorofthedenominatorproducesapartialfraction.Afactorax+bproduces
a partialfraction of the form A whereAisan unknown constant.
ax+b
(3) Evaluate the unknown constants of the partial fractions. This is done by evaluation
using a specific value ofxor by equating coefficients.
A linear factor ax +bin the denominator produces a partial fraction of the form
A
ax+b .
Example1.38 Express
6x+8
x 2 +3x+2
as itspartialfractions.
Solution The denominator isfactorized as
x 2 +3x+2=(x+1)(x+2)