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

494 Chapter 7 Techniques of Integration
In the case of an Equation 1 whose denominator is more complicated, the next step
is to factor the denominator Qsxd as far as possible. It can be shown that any polynomial
Q can be factored as a product of linear factors (of the form ax 1 bd and irreducible
quadratic factors (of the form ax 2 1 bx 1 c, where b 2 2 4ac , 0). For instance, if
Qsxd − x 4 2 16, we could factor it as
Qsxd − sx 2 2 4dsx 2 1 4d − sx 2 2dsx 1 2dsx 2 1 4d
The third step is to express the proper rational function RsxdyQsxd (from Equation 1)
as a sum of partial fractions of the form
A
sax 1 bd or Ax 1 B
i sax 2 1 bx 1 cd j
A theorem in algebra guarantees that it is always possible to do this. We explain the
details for the four cases that occur.
Case I The denominator Qsxd is a product of distinct linear factors.
This means that we can write
Qsxd − sa 1 x 1 b 1 dsa 2 x 1 b 2 d ∙ ∙ ∙ sa k x 1 b k d
where no factor is repeated (and no factor is a constant multiple of another). In this case
the partial fraction theorem states that there exist constants A 1 , A 2 , . . . , A k such that
2
Rsxd
Qsxd − A 1
a 1 x 1 b 1
1
A 2
a 2 x 1 b 2
1 ∙ ∙ ∙ 1
A k
a k x 1 b k
These constants can be determined as in the following example.
Example 2 Evaluate y
x 2 1 2x 2 1
2x 3 1 3x 2 2 2x dx.
SOLUTION Since the degree of the numerator is less than the degree of the denominator,
we don’t need to divide. We factor the denominator as
2x 3 1 3x 2 2 2x − xs2x 2 1 3x 2 2d − xs2x 2 1dsx 1 2d
Since the denominator has three distinct linear factors, the partial fraction decomposition
of the integrand (2) has the form
3
x 2 1 2x 2 1
xs2x 2 1dsx 1 2d − A x 1
B
2x 2 1 1
C
x 1 2
Another method for finding A, B, and C
is given in the note after this example.
To determine the values of A, B, and C, we multiply both sides of this equation by the
product of the denominators, xs2x 2 1dsx 1 2d, obtaining
4
x 2 1 2x 2 1 − As2x 2 1dsx 1 2d 1 Bxsx 1 2d 1 Cxs2x 2 1d
Expanding the right side of Equation 4 and writing it in the standard form for polyno
mials, we get
5
x 2 1 2x 2 1 − s2A 1 B 1 2Cdx 2 1 s3A 1 2B 2 Cdx 2 2A
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