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

496 Chapter 7 Techniques of Integration
Since ln x 2 ln y − lnsxyyd, we can write the integral as
6
y
dx
x 2 2 a 2
− 1 2a ln Z
x 2 a
x 1 a
Z 1 C
See Exercises 57–58 for ways of using Formula 6.
n
Case II Qsxd is a product of linear factors, some of which are repeated.
Suppose the first linear factor sa 1 x 1 b 1 d is repeated r times; that is, sa 1 x 1 b 1 d r occurs
in the factorization of Qsxd. Then instead of the single term A 1 ysa 1 x 1 b 1 d in Equation
2, we would use
7
A 1
a 1 x 1 b 1
1
A 2
sa 1 x 1 b 1 d 2 1 ∙ ∙ ∙ 1
A r
sa 1 x 1 b 1 d r
By way of illustration, we could write
x 3 2 x 1 1
− A x 2 sx 2 1d 3 x 1 B x 1 C
2 x 2 1 1
but we prefer to work out in detail a simpler example.
Example 4 Find y x 4 2 2x 2 1 4x 1 1
x 3 2 x 2 2 x 1 1
dx.
D
sx 2 1d 1 E
2 sx 2 1d 3
SOLUTION The first step is to divide. The result of long division is
x 4 2 2x 2 1 4x 1 1
x 3 2 x 2 2 x 1 1
− x 1 1 1
4x
x 3 2 x 2 2 x 1 1
The second step is to factor the denominator Qsxd − x 3 2 x 2 2 x 1 1. Since
Qs1d − 0, we know that x 2 1 is a factor and we obtain
x 3 2 x 2 2 x 1 1 − sx 2 1dsx 2 2 1d − sx 2 1dsx 2 1dsx 1 1d
− sx 2 1d 2 sx 1 1d
Since the linear factor x 2 1 occurs twice, the partial fraction decomposition is
4x
sx 2 1d 2 sx 1 1d −
A
x 2 1 1
B
sx 2 1d 1 C
2 x 1 1
Multiplying by the least common denominator, sx 2 1d 2 sx 1 1d, we get
8
4x − Asx 2 1dsx 1 1d 1 Bsx 1 1d 1 Csx 2 1d 2
− sA 1 Cdx 2 1 sB 2 2Cdx 1 s2A 1 B 1 Cd
Another method for finding the
coefficients:
Put x − 1 in (8): B − 2.
Put x − 21: C − 21.
Put x − 0: A − B 1 C − 1.
Now we equate coefficients:
A 1 C − 0
B 2 2C − 4
2A 1 B 1 C − 0
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