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Section 8.5 – Partial Fractions
White Board Challenge
Combine the two fractions into a single fraction:
5 10
4 1
x
x x
2
5 10
3 4
xor
x x
2 3 2 1 3 4
4 1 4 1 1 4
x x
x x x x x x
Find a
common denominator:
2 2 3 12
4 1 4 1
x x
x x x x
White Board Challenge
Reverse the process and write as the sum of two fractions in the form: .
2
1 2
Ax Bx A B
x x
2
2
2
A B x A B
x x
2 1
1 2 1 2 2 1
A B A x B x
x x x x x x
Use the final form to start:
2
1 2 1 2
Ax A Bx B
x x x x
2 2
2 1 5
2 2
A B x A B x
x x x x
Since:
1A B 2 5A B _________
3 6A2A
2 1B 1B
Write and solve a
System of Equations:
2 1
1 2x x
Write the
final answer:
Integration of Rational FractionsWe can integrate some rational functions by expressing it as a sum of simpler fractions, called partial fractions, that we already know how to integrate. (We can assume that the degree of the numerator is smaller than the degree of the denominator. Otherwise we could divide and work with the remainder.)
From Algebra we can find a common denominator to show:
Thus, if we reverse the procedure, we see how to integrate the function on the right side of the equation:
Use this method if no previous technique works and the denominator factors nicely.
2
2 1 5
1 2 2
x
x x x x
2
5 2 12ln 1 ln 2
2 1 2
xdx dx x x C
x x x x
Example 1Evaluate
2
1
5 6 2 3
dxdx
x x x x
Factor the
denominator first:
Use Partial Fractions to rewrite the integral:
3 2
2 3
Ax Bx A B
x x
2
3 2
5 6
A B x A B
x x
3 2
2 3 2 3 3 2
A B A x B x
x x x x x x
3 2
2 3 2 3
Ax A Bx B
x x x x
2 2
3 2 0 1
5 6 5 6
A B x A B x
x x x x
0A B
3 2 1A B or B A3 2 1A A
1A1B
Write and solve a
System of Equations:
Integrate:
2 3
A Bdx
x x
1 1
2 3dx
x x ln 2 ln 3x x C
Since:
Example 2Evaluate
2 2
3 2
2 1 2 1
2 3 2 2 1 2
x x x xdx dx
x x x x x x
Factor the denominator first:
Use Partial Fractions to rewrite the integral:
2 1 2
A B Cdx
x x x
22 1 2 2 2 1 2 1A x x Bx x Cx x x x 2 2 2 22 3 2 2 2 2 1Ax Ax A Bx Bx Cx Cx x x
2 22 2 3 2 2 1 2 1A B C x A B C x A x x
2 2 1A B C 3 2 2A B C
2 1A
Write and solve a
System of Equations:
12
15
110
A
B
C
Trick: Multiple each numerator by the other denominators and set it equal to the original numerator:
Distribute and combine
like terms
Example 2 (continued)Evaluate
2 2
3 2
2 1 2 1
2 3 2 2 1 2
x x x xdx dx
x x x x x x
Factor the denominator first:
Use Partial Fractions to rewrite the integral:
2 1 2
A B Cdx
x x x
1 115 102
2 1 2dx
x x x
Integrate: 1 1 1
2 5 10
1 1 1
2 1 2dx dx dxx x x
1 1 12 10 10
1 2 1
2 1 2dx dx dxx x x
1 1 12 10 10ln ln 2 1 ln 2x x x C
Since:
12
15
110
A
B
C
Example 2 EASIEREvaluate
2 2
3 2
2 1 2 1
2 3 2 2 1 2
x x x xdx dx
x x x x x x
Factor the denominator first:
Use Partial Fractions to rewrite the integral:
2 1 2
A B Cdx
x x x
20 2 0 1
0 2 0 1 0 2A
Trick: Ignore the Discontinuity.
1
1 2
1
2The “A” fraction
is undefined at x=0.
Substitute x=0 into the original factored
integral to find A.
20.5 2 0.5 1
0.5 0.5 22 0 1B
0.25
0.5 2.5 1
5The “B” fraction
is undefined at x=1/2.
Substitute x=1/2 into the original factored
integral to find B.
22 2 2 1
2 2 2 22 1C
1
2 5
1
10The “C” fraction
is undefined at x=-2.
Substitute x=-2 into the original factored
integral to find C.
IGNORE
IGNORE
IGNORE
Now find the antiderivative.
Example 3A deer population grows logistically with growth constant in a forest with a carrying capacity of 1000 deer. If the initial population is 100 deer, find the particular solution that models the population.
10000.4 1dP Pdt P
Substitute into the Logistic Model:
Separate the variables :
1000
11
0.4PPdP dt
1000
11
0.4PPdP dt
10001
0.4PA BP dP dt
Use Partial Fractions:
01000
10 1
A
Igno
re t
he
Dis
cont
inui
ty:
1
10001000
11000 1
B
0.001--
-----
Find the antiderivative:
1000
0.00111
0.4PP dP dt 1 1
1000 0.4P P dP dt ln ln 1000 0.4P P t C
1000ln 0.4PP t C
0.41000
t CPP e
0.4
1000C tP
P e e 0.4
1000tP
P Ce
Use the initial condition:
0.4 01001000 100 Ce
19C
Solve for P:0.41
1000 9tP
P e 0.49 1000 tP P e 0.49 1000tPe P
0.49 1 1000tP e
0.4
1000
1 9 tP t
e
Most Logistic Model AP exam questions can be answered without solving the differential equation analytically. But seeing it can’t hurt.
White Board Challenge
Evaluate:
1
2
xdx
x x
1 12 2ln ln 2x x C
