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

498 Chapter 7 Techniques of Integration
In order to integrate the second term we split it into two parts:
y x 2 1
x 2 1 4 dx − y
x
x 2 1 4 dx 2 y
1
x 2 1 4 dx
We make the substitution u − x 2 1 4 in the first of these integrals so that du − 2x dx.
We evaluate the second integral by means of Formula 10 with a − 2:
y 2x 2 2 x 1 4
xsx 2 1 4d
dx − y 1 x dx 1 y
x
x 2 1 4 dx 2 y
1
x 2 1 4 dx
− ln | x | 1 1 2 lnsx 2 1 4d 2 1 2 tan21 sxy2d 1 K
n
Example 6 Evaluate y 4x 2 2 3x 1 2
4x 2 2 4x 1 3 dx.
SOLUTION Since the degree of the numerator is not less than the degree of the denominator,
we first divide and obtain
4x 2 2 3x 1 2
4x 2 2 4x 1 3 − 1 1 x 2 1
4x 2 2 4x 1 3
Notice that the quadratic 4x 2 2 4x 1 3 is irreducible because its discriminant is
b 2 2 4ac − 232 , 0. This means it can’t be factored, so we don’t need to use the
partial fraction technique.
To integrate the given function we complete the square in the denominator:
4x 2 2 4x 1 3 − s2x 2 1d 2 1 2
This suggests that we make the substitution u − 2x 2 1. Then du − 2 dx and
x − 1 2su 1 1d, so
y 4x 2 2 3x 1 2
4x 2 2 4x 1 3 dx − y S1 1
− x 1 1 2 y
− x 1 1 4 y
x 2 1
4x 2 2 4x 1 3D dx
1
2 su 1 1d 2 1
du − x 1 1
u 2 4 y u 2 1
1 2
u 2 1 2 du
u
u 2 1 2 du 2 1 4 y
1
u 2 1 2 du
21S − x 1 1 8 lnsu 2 1 2d 2 1 4 ? 1
tan D u 1 C
s2 s2
− x 1 1 8 lns4x 2 2 4x 1 3d 2 1 2x 2 1
tan21S D 1 C n
4s2 s2
NOTE Example 6 illustrates the general procedure for integrating a partial fraction
of the form
Ax 1 B
where b 2 2 4ac , 0
ax 2 1 bx 1 c
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