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

1.7 Partial fractions 41
Thus we have two simultaneous equations in A and B, which may be solved to give
A = 2 andB = 4 as before.
1.7.2 Repeatedlinearfactor
We now examine proper fractions whose denominators factorize into linear factors,
where one or more of the linear factors isrepeated.
Arepeated linear factor, (ax +b) 2 , produces two partialfractions of the form
A
ax+b +
B
(ax +b) 2
Arepeated linear factor, (ax +b) 2 , leadstopartialfractions
A
ax+b +
B
(ax +b) 2
Example1.39 Express
2x+5
x 2 +2x+1
as partialfractions.
Solution Thedenominatorisfactorizedtogive (x+1) 2 .Herewehaveacaseofarepeatedfactor.
A
This repeated factor generates partialfractions
x +1 + B
(x +1) 2.
Thus
2x+5
x 2 +2x+1 = 2x+5
(x +1) 2 = A
x +1 +
Multiplying by (x +1) 2 gives
2x+5=A(x+1)+B=Ax+A+B
B
(x +1) 2
Equating coefficients ofxgivesA = 2.Evaluation withx = −1 givesB = 3.So
2x+5
x 2 +2x+1 = 2
x +1 + 3
(x +1) 2
Example1.40 Express
14x 2 +13x
(4x 2 +4x +1)(x −1)
as partialfractions.
Solution Thedenominatorisfactorizedto (2x+1) 2 (x−1).Therepeatedfactor, (2x+1) 2 ,produces
partialfractions of the form
A
2x+1 + B
(2x +1) 2