Accumulation
Authors: Bill Davis, Horacio Porta and Jerry Uhl
©19962007
Publisher: Math Everywhere, Inc.
Version 6.0
2.07 Traditional Pat Integration Procedures
BASICS
Traditionally, calculus courses spend a lot of time and effort on practicing the art
of evaluating integrals. By now you will have noticed that such practice isn't a
central issue in Calculus&
Mathematica
. It is important to understand the skills,
though. Here's a compendium of techniques.
B.1) Partial fractions for quotients of polynomials
·
B.1.a)
Look at
Mathematica
's calculation of
Ÿ
2.1
5.3
1
x
2

1
„
x
Clear
@
f, x
D
;
f
@
x
_
D
=
1
x
2

1
;
‡
2.1
5.3
f
@
x
D
„
x
0.327079
Explain where this calculation comes from.
·
Answer:
Look at the function you are integrating:
Clear
@
f, x
D
;
f
@
x
_
D
=
1
x
2

1
1

1
+
x
2
Factor the denominator of
f
@
x
D
:
Factor
@
Denominator
@
f
@
x
DDD
H

1
+
x
L H
1
+
x
L
Put:
Clear
@
newf, a, b
D
;
newf
@
x
_
D
=
a
1
+
x
+
b

1
+
x
b

1
+
x
+
a
1
+
x
Now put
newf
@
x
D
over a common denominator:
ExpandAll
@
Together
@
newf
@
x
DDD

a

1
+
x
2
+
b

1
+
x
2
+
ax

1
+
x
2
+
bx

1
+
x
2
You want this to be:
f
@
x
D
1

1
+
x
2
You can arrange this by solving for the
a and
b that make:
eqn1
=
H

a
+
b
==
1
L
;
eqn2
=
H
a
+
b
==
0
L
;
Solve
@8
eqn1, eqn2
<D
::
a
Ø 
1
2
, b
Ø
1
2
>>
This tells you that when you go with
a
= 
1
2
and
b
=
1
2
, then
newf
@
x
D
=
f
@
x
D
.
a
= 
1
2
;
b
=
1
2
;
newf
@
x
D
1
2
H

1
+
x
L

1
2
H
1
+
x
L
The point of this is that
Ÿ
2.1
5.3
newf
@
x
D
„
x
is easy to integrate by hand.
It is
1
2
Log
@

1
+
x
D

1
2
Log
@
1
+
x
D
»
2.1
5.3
:
top
=
1
2
Log
@

1
+
x
D

1
2
Log
@
1
+
x
D
ê
.x
>
5.3;
bottom
=
1
2
Log
@

1
+
x
D

1
2
Log
@
1
+
x
D
ê
.x
>
2.1;
integral
=
top

bottom
0.327079
Check:
‡
2.1
5.3
f
@
x
D
„
x
0.327079
Got it.
·
B.1.b)
Look at
Mathematica
's evaluation of
Ÿ
0
t
4
x
+
1
H
x
+
2
L H
x
+
1
L
2
„
x:
Clear
@
x, t
D
;
‡
0
t
4x
+
1
H
x
+
2
L H
x
+
1
L
2
„
x
If
B
Re
@
t
D
¥ 
1
»»
Im
@
t
D
!
0,

3t
1
+
t
+
Log
@
128
D
+
7Log
@
1
+
t
D

7Log
@
2
+
t
D
,
Integrate
B
1
H
1
+
x
L
2
H
2
+
x
L
+
4x
H
1
+
x
L
2
H
2
+
x
L
,
8
x, 0, t
<
, Assumptions
Ø !
H
Re
@
t
D
¥ 
1
»»
Im
@
t
D
!
0
LFF
Explain where this calculation comes from.
·
Answer:
Look at the function you are integrating:
Clear
@
f, x
D
;
f
@
x
_
D
=
4x
+
1
H
x
+
2
L H
x
+
1
L
2
1
+
4x
H
1
+
x
L
2
H
2
+
x
L
Factor the denominator of
f
@
x
D
:
Factor
@
Denominator
@
f
@
x
DDD
H
1
+
x
L
2
H
2
+
x
L
Put:
Clear
@
newf, a, b, c
D
;
newf
@
x
_
D
=
a
1
+
x
+
b
H
1
+
x
L
2
+
c
2
+
x
b
H
1
+
x
L
2
+
a
1
+
x
+
c
2
+
x
Now put
newf
@
x
D
over a common denominator:
Together
@
newf
@
x
DD
2a
+
2b
+
c
+
3ax
+
bx
+
2cx
+
ax
2
+
cx
2
H
1
+
x
L
2
H
2
+
x
L
You want this to be:
f
@
x
D
1
+
4x
H
1
+
x
L
2
H
2
+
x
L
You can arrange this by solving for the
a,
b and
c that make:
eqn1
=
H
2a
+
2b
+
c
==
1
L
;
eqn2
=
H
3a
+
b
+
2c
==
4
L
;
eqn3
=
H
a
+
c
==
0
L
;
Solve
@8
eqn1, eqn2, eqn3
<D
88
a
Ø
7, c
Ø 
7, b
Ø 
3
<<
This tells you that when you go with
a
=
7,
b
= 
3 and
c
= 
7 then
newf
@
x
D
=
f
@
x
D
.
a
=
7;
b
= 
3;
c
= 
7;
newf
@
x
D

3
H
1
+
x
L
2
+
7
1
+
x

7
2
+
x
Confirm:
Apart
@
f
@
x
DD

3
H
1
+
x
L
2
+
7
1
+
x

7
2
+
x
The point of this is that
Ÿ
0
t
newf
@
x
D
„
x
is easy to integrate by hand.
It is
3
1
+
x
+
7 Log
@
1
+
x
D

7 Log
@
2
+
x
D
»
0
t
:
top
=
3
1
+
x
+
7Log
@
1
+
x
D

7Log
@
2
+
x
D ê
.x
Ø
t;
bottom
=
3
1
+
x
+
7Log
@
1
+
x
D

7Log
@
2
+
x
D ê
.x
Ø
0;
integral
=
top

bottom

3
+
3
1
+
t
+
7Log
@
2
D
+
7Log
@
1
+
t
D

7Log
@
2
+
t
D
This is equivalent to the
Mathematica
calculation:
Clear
@
x, t
D
;
‡
0
t
4x
+
1
H
x
+
2
L H
x
+
1
L
2
„
x
If
B
Re
@
t
D
¥ 
1
»»
Im
@
t
D
!
0,

3t
1
+
t
+
Log
@
128
D
+
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 Spring '08
 Staff
 Calculus, Sin, td, Mathematica

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