E
INTEGRATION
STRATEGIES
T
he set of exercises in this section consists of an assortment of indefinite integrals that
can be evaluated using the techniques developed in Chapter 7. Your task is to decide
which techniques are best suited to the given integral and to carry out the computation.
Let’s review the tools at our disposal. The two most powerful tools are substitution
and integration by parts. In addition, we have special methods for treating specific types
of integrals:
(A)
Trigonometric integrals:
use trigonometric identities and reduction formulas (Sec
tion 7.3)
(B)
Square root expressions
√
x
2
±
a
2
and
√
a
2
−
x
2
:
use trigonometric substitution
(Section 7.4) or hyperbolic substitution (Section 7.5)
(C)
Rational functions:
use the method of partial fractions (Section 7.6)
When confronted with an integration problem, you may ask yourself the following ques
tions:
1.
Can the integral be evaluated directly (perhaps after algebraic manipulation or a sub
stitution)?
2.
Is the integral amenable to integration by parts (or perhaps substitution followed by
integration by parts)?
3.
Does the integral fall into one of the types (A)–(C) described above? Or can it be
reduced to one of these types by a substitution?
You may also refer to the Table of Integrals at the back of the book. This table
contains a list of standard integration formulas, including reduction formulas for trigono
metric, exponential, and inverse trigonometric integrals. Sometimes it is necessary to
transform your integral before applying one of the formulas listed in the table. In all
cases, be on the lookout for the simplest method.
EXAMPLE 1
Look for the Simplest Method
Evaluate the integrals
(a)
x
7
dx
x
4
−
1
and
(b)
x
2
dx
x
4
−
1
.
Solution
At first glance, both integrals appear to be candidates for the method of partial
fractions. However, the first integral may be treated much more simply using the substi
tution
u
=
x
4
−
1.
E1
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E2
A P P E N D I X
E
INTEGRATION STRATEGIES
(a)
Let
u
=
x
4
−
1 and
du
=
4
x
3
dx
. Then
x
4
=
u
+
1 and
x
7
dx
=
x
4
x
3
dx
=
1
4
(
u
+
1
)
du
1
x
7
dx
x
4
−
1
=
1
4
(
u
+
1
)
du
u
=
1
4
1
+
1
u
du
(
simplify algebraically
)
=
1
4
u
+
1
4
ln

u
 +
C
=
1
4
(
x
4
−
1
)
+
1
4
ln

x
4
−
1
 +
C
We may absorb the term
−
1
4
into the constant
C
and write our result as
x
7
dx
x
4
−
1
=
1
4
x
4
+
1
4
ln

x
4
−
1
 +
C
(b)
Direct substitution cannot be used to evaluate the second integral. We must find the
The substitution
u
=
x
4
−
1
is not effective
in integral (b) of Example 1 because
x
2
dx
=
x
−
1
x
3
dx
=
1
4
(
u
+
1
)
−
1
/
4
du
The substitution does not lead to a simpler
integral:
x
2
dx
x
4
−
1
=
1
4
(
u
+
1
)
−
1
/
4
du
u
partial fraction decomposition using the factorization
x
4
−
1
=
(
x
−
1
)(
x
+
1
)(
x
2
+
1
)
:
x
2
x
4
−
1
=
1
4
x
−
1
−
1
4
x
+
1
+
1
2
x
2
+
1
x
2
dx
x
4
−
1
=
1
4
dx
x
−
1
−
1
4
dx
x
+
1
+
1
2
dx
x
2
+
1
=
1
4
ln

x
−
1
 −
1
4
ln

x
+
1
 +
1
2
tan
−
1
x
+
C
EXAMPLE 2
Find a Simplifying Substitution
Find approaches to evaluate
(a)
e
x
dx
√
1
−
e
2
x
and
(b)
e
2
x
dx
√
1
−
e
2
x
.
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 Spring '08
 ningzhong
 Calculus, Integrals, dx

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