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

44 Chapter 1 Review of algebraic techniques
Letthedegreeofthenumeratorbenandthedegreeofthedenominatorbed.Ifn d
thenthefractionisimproper.Improperfractionshavepartialfractionsinadditionto
thosegeneratedbythefactorsofthedenominator.Theseadditionalpartialfractions
take the form ofapolynomial of degreen−d.
Example1.42 Express as partialfractions
4x 3 +10x+4
2x 2 +x
Solution Thedegreeofthenumeratoris3,thatisn = 3.Thedegreeofthedenominatoris2,that
isd = 2.Thus,the fraction isimproper.
Nown −d = 1andthisisameasureoftheextenttowhichthefractionisimproper.
Thepartialfractionswillincludeapolynomialofdegree1,thatisAx +B,inadditionto
the partialfractions generated by the factors of the denominator.
The denominator factorizes tox(2x +1). These factors generate partial fractions of
the form C x + D
2x+1 . Hence
4x 3 +10x+4
2x 2 +x
= 4x3 +10x+4
x(2x +1)
Multiplying byxand 2x +1 yields
=Ax+B+ C x +
D
2x+1
4x 3 +10x +4 = (Ax +B)x(2x +1) +C(2x +1) +Dx (1.12)
The constantsA,B,C andDcan now beevaluated.
Puttingx = 0 into Equation (1.12) gives
4 =C
Puttingx = −0.5 into Equation (1.12) gives
−1.5 = − D 2
D = 3
Equating coefficients ofx 3 gives
4=2A
A = 2
Equating coefficients ofxgives
Hence
10=B+2C+D
B=−1
4x 3 +10x+4
2x 2 +x
=2x−1+ 4 x + 3
2x+1