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The Empire Builder, 1957
6.5 day 1 Partial Fractions
Greg Kelly, Hanford High School, Richland, Washington
2
5 3
2 3
xdx
x x
This would be a lot easier if we could
re-write it as two separate terms.
5 3
3 1
x
x x
3 1
A B
x x
1
These are called non-repeating linear factors.
You may already know a short-cut for this type of problem. We will get to that in a few minutes.
2
5 3
2 3
xdx
x x
This would be a lot easier if we could
re-write it as two separate terms.
5 3
3 1
x
x x
3 1
A B
x x
Multiply by the common denominator.
5 3 1 3x A x B x
5 3 3x Ax A Bx B Set like-terms equal to each other.
5x Ax Bx 3 3A B
5 A B 3 3A B Solve two equations with two unknowns.
1
2
5 3
2 3
xdx
x x
5 3
3 1
x
x x
3 1
A B
x x
5 3 1 3x A x B x
5 3 3x Ax A Bx B
5x Ax Bx 3 3A B
5 A B 3 3A B Solve two equations with two unknowns.
5 A B 3 3A B
3 3A B
8 4B
2 B 5 2A
3 A
3 2
3 1dx
x x
3ln 3 2ln 1x x C
This technique is calledPartial Fractions
1
2
5 3
2 3
xdx
x x
The short-cut for this type of problem is
called the Heaviside Method, after English engineer Oliver Heaviside.
5 3
3 1
x
x x
3 1
A B
x x
Multiply by the common denominator.
5 3 1 3x A x B x
8 0 4A B
1
Let x = - 1
2 B
12 4 0A B Let x = 3
3 A
2
5 3
2 3
xdx
x x
The short-cut for this type of problem is
called the Heaviside Method, after English engineer Oliver Heaviside.
5 3
3 1
x
x x
3 1
A B
x x
5 3 1 3x A x B x
8 0 4A B
1
2 B
12 4 0A B 3 A
3 2
3 1dx
x x
3ln 3 2ln 1x x C
Good News!
The AP Exam only requires non-repeating linear factors!
The more complicated methods of partial fractions are good to know, and you might see them in college, but they will not be on the AP exam or on my exam.
2
6 7
2
x
x
Repeated roots: we must use two terms for partial fractions.
22 2
A B
x x
6 7 2x A x B
6 7 2x Ax A B
6x Ax 7 2A B
6 A 7 2 6 B
7 12 B
5 B
2
6 5
2 2x x
2
3 2
2
2 4 3
2 3
x x x
x x
If the degree of the numerator is higher than the degree of the denominator, use long division first.
2 3 22 3 2 4 3x x x x x 2x
3 22 4 6x x x 5 3x
2
5 32
2 3
xxx x
5 3
23 1
xx
x x
3 2
23 1
xx x
4
(from example one)
22
2 4
1 1
x
x x
irreduciblequadratic
factor
repeated root
22 1 1 1
Ax B C D
x x x
first degree numerator
2 2 22 4 1 1 1 1x Ax B x C x x D x
2 3 2 22 4 2 1 1x Ax B x x C x x x Dx D
3 2 2 3 2 22 4 2 2x Ax Ax Ax Bx Bx B Cx Cx Cx C Dx D
A challenging example:
3 2 2 3 2 22 4 2 2x Ax Ax Ax Bx Bx B Cx Cx Cx C Dx D
0 A C 0 2A B C D 2 2A B C 4 B C D
1 0 1 0 0
2 1 1 1 0
1 2 1 0 2
0 1 1 1 4
2 r 3
r 1
1 0 1 0 0
0 3 1 1 4
0 2 0 0 2
0 1 1 1 4
2
1 0 1 0 0
0 1 0 0 1
0 3 1 1 4
0 1 1 1 4
3 r 2
r 2
1 0 1 0 0
0 1 0 0 1
0 0 1 1 1
0 0 1 1 3
r 3
1 0 1 0 0
0 1 0 0 1
0 3 1 1 4
0 1 1 1 4
3 r 2
r 2
1 0 1 0 0
0 1 0 0 1
0 0 1 1 1
0 0 1 1 3
r 3
1 0 1 0 0
0 1 0 0 1
0 0 1 1 1
0 0 0 2 2
2
1 0 1 0 0
0 1 0 0 1
0 0 1 1 1
0 0 0 1 1
r 4
1 0 1 0 0
0 1 0 0 1
0 0 1 0 2
0 0 0 1 1
r 3
1 0 0 0 2
0 1 0 0 1
0 0 1 0 2
0 0 0 1 1
1 0 0 0 2
0 1 0 0 1
0 0 1 0 2
0 0 0 1 1
22
2 4
1 1
x
x x
22 1 1 1
Ax B C D
x x x
22
2 1 2 1
1 1 1
x
x x x
We can do this problem on the TI-89:
22
2 4expand
1 1
x
x x
expand ((-2x+4)/((x^2+1)*(x-1)^2))
22 2
2 1 2 1
1 1 1 1
x
x x x x
Of course with the TI-89, we could just integrate and wouldn’t need partial fractions!
3F2
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