Calculus: Analytic Geometry and Calculus, with Vectors

512 Exponential and logarithmic functions

of the numerator is less than that of the denominator. Letting f(x) be
defined by (9.413), we divide to obtain

(9.42) f (x) = x2 - 2x -I- 1 -I-x

3x

3 - (^22) + x
Our difficulties will have been surmounted when we succeed in expressing
the last quotient (or fraction) in (9.42) as a sum of simpler quotients (or
fractions) that are called partial fractions. This brings us to the key
problem of this section, namely, the problem of representing a given
proper rational function (that is, a quotient of polynomials in which the
degree of the numerator is less than that of the denominator) in terms of
partial fractions. This key problem is important outside as well as inside
the calculus. The problem is treated in some algebra books, but students
normally make their first acquaintance with the problem in calculus
books.
To begin discovery of the partial fractions whose sum is a given proper
rational fraction, we must factor the given denominator and use these
factors to determine the nature of the partial fractions. Electronic
computers are often used to factor denominators, but factoring of the
denominator of the quotient in (9.42) is quite easy if we happen to notice
that

xs - x2 + x - 1 = (x - 1)x2 + (x - 1) = (x - 1)(x2 + 1).

Thus the quotient in (9.42) is the left member of the equality

3x2+1 __ 11 Bx+C

(9.43)

(x - 1) (x2 T 1) x - 1+ x2+ 1

While we are entitled to be quite mystified by the fact until the matter

has been investigated, it is possible to determine three constants .1, B, C

such that (9.43) is true for each x for which the denominators are all

different from zero. In fact, we can clear the denominators from (9.43)

and determine the constants so that the formula

(9.431) 3x2 + 1 = 1(x2 + 1) + (Bx + C)(x - 1)

is true for each x. One way to prove this and to find the constants is to

put (9.431) in the form

3x2 + 1 = (fl + B)x2 + (-B + C)x + (4 - C)

and to observe that this is surely an identity in x if

A+B=3, -B+C=O, Z-C=1

and hence (as we show by solving these equations) if

(9.432) Z = 2, B = 1, C = 1.