Calculus: Analytic Geometry and Calculus, with Vectors

9.4 Partial fractions 513

There is, however, an easier way to find 11, B, C directly from (9.431).
Putting x = 1 in (9.431) shows immediately that 4 = A(2) and hence
that 4=2. Putting x = 0 shows that 1 = A - C or 1 =2-C,
so C = 1. Finally, putting x = 2 shows that 13= 2.5 + (2B + 1),
so B = 1. Substituting from (9.432) into (9.43) gives

(9.433)

3x2 + 1 - 2 x + 1

( X X 2

Since this is the last quotient in (9.42), we obtain (9.411). Thus we

have succeeded in writing the given quotient in (9.413) as the sum of a
polynomial and partial fractions.
Some problems are easier than the one we have solved, andsome are
more difficult. When we wish to integrate the left member of

2x+6 .4 B

(9.44) (x

• 1)(x + 1) x - 1+ -x+ 1'

we determine -4 and B so that (9.44) and

(9.441) 2x + 6 = A(x + 1) + B(x - 1)

hold. Putting x = 1 shows that -4 = 4, and putting x = -1 shows

that B = -2. Thus

2x+6 4 _ 2

(x - 1)(x + 1) x - 1 x + 1'

and integration gives

f2?+dx-1 = 4log x-1- 2log'x l 11 + c.

The result can, for better or for worse, be put in the form

f

2x

+ 6dx= log(x 1)2 + c.

The problems at the end of this section provide additional clues to
methods by which quotients are expressed in terms of partial fractions.
Problem 11 proposes study of basic theory, and persons interested in
more than the simplest mechanical aspects of our subject can do this
studying at any time and perhaps even more than once.

Problems 9.49

1 Show that, when p 0 q and a> 0,

(1) f (^1) dx = 1 loglx- pl+c.
a(x - p)(x - q) a(p - q) x - q