James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 7.4 Integration of Rational Functions by Partial Fractions 493
In this section we show how to integrate any rational function (a ratio of polynomials) by
expressing it as a sum of simpler fractions, called partial fractions, that we already know
how to integrate. To illustrate the method, observe that by taking the fractions 2ysx 2 1d
and 1ysx 1 2d to a common denominator we obtain
2
x 2 1 2 1 2sx 1 2d 2 sx 2 1d
− − x 1 5
x 1 2 sx 2 1dsx 1 2d x 2 1 x 2 2
If we now reverse the procedure, we see how to integrate the function on the right side
of this equation:
y
x 1 5
x 2 1 x 2 2 dx − y S 2
x 2 1 2 1 dx
x 1 2D
− 2 ln | x 2 1 | 2 ln | x 1 2 | 1 C
To see how the method of partial fractions works in general, let’s consider a rational
function
f sxd − Psxd
Qsxd
where P and Q are polynomials. It’s possible to express f as a sum of simpler fractions
pro vided that the degree of P is less than the degree of Q. Such a rational function
is called proper. Recall that if
Psxd − a n x n 1 a n21 x n21 1 ∙ ∙ ∙ 1 a 1 x 1 a 0
where a n ± 0, then the degree of P is n and we write degsPd − n.
If f is improper, that is, degsPd > degsQd, then we must take the preliminary step
of dividing Q into P (by long division) until a remainder Rsxd is obtained such that
degsRd , degsQd. The division statement is
1
f sxd − Psxd Rsxd
− Ssxd 1
Qsxd Qsxd
where S and R are also polynomials.
As the following example illustrates, sometimes this preliminary step is all that is
required.
≈+x +2
x-1)˛ +x
˛-≈
≈+x
≈-x
2x
2x-2
2
Example 1 Find y x 3 1 x
x 2 1 dx.
SOLUTION Since the degree of the numerator is greater than the degree of the denominator,
we first perform the long division. This enables us to write
y x 3 1 x
x 2 1 dx − y Sx 2 1 x 1 2 1 2
x 2 1D dx
− x 3
3 1 x 2
2 1 2x 1 2 ln | x 2 1 | 1 C n
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