Partial Fraction Decomposition
Breaking down and integrating
rational functions
Integrate the following function…..
1
1
−
𝑑𝑥
𝑥−3 𝑥−2
Now what if we disguise this equation
a bit by combining……
1
1
−
𝑑𝑥 =
𝑥−3 𝑥−2
𝑥−2 −𝑥+3
𝑑𝑥 =
2
𝑥 − 5𝑥 + 6
1 𝑥−2 −1 𝑥−3
𝑑𝑥
(𝑥 − 3)(𝑥 − 2)
1
𝑑𝑥
2
𝑥 − 5𝑥 + 6
If we are given this equation initially:
1
𝑑𝑥
2
𝑥 −5𝑥+6
We are going to have to do some expansion in order
to put it into a form that is easy to integrate.
This process is called PARTIAL FRACTION DECOMPOSITION.
*
*
Now you want to get a common denominator in
order to set the numerators equal to one another.
The method of Partial Fraction
Decomposition ALWAYS works when you
are integrating a rational function.
Rational Function = Ratio of polynomials
You will decompose/expand the rational
function so it can be easily integrated.
1
𝐴
𝐵
=
+
2
𝑥 −1 𝑥+1 𝑥−1
1 = 𝐴 𝑥 − 1 + 𝐵(𝑥 + 1)
Let 𝑥 = −1
1
𝐴=−
2
Let 𝑥 = 1
1
𝐵=
2
𝟏
𝟏
−
𝟐 + 𝟐
𝒅𝒙
𝒙+𝟏 𝒙−𝟏
𝟏
𝟏
𝟏
𝒅𝒙 = − 𝐥𝐧 𝒙 + 𝟏 + 𝐥𝐧 𝒙 − 𝟏 + 𝑪
𝟐
𝒙 −𝟏
𝟐
𝟐
1
𝐴
𝐵
=
+
2
4𝑥 − 9 2𝑥 + 3 2𝑥 − 3
1 = 𝐴 2𝑥 − 3 + 𝐵(2𝑥 + 3)
Let 𝑥 =
3
2
1 = 6𝐵
1
𝐵=
6
Let 𝑥 = −
3
2
1 = −6𝐴
1
−1/6
1/6
=
+
2
4𝑥 − 9 2𝑥 + 3 2𝑥 − 3
1
1
− ln 2𝑥 + 3 + ln 2𝑥 − 3 + 𝐶
12
12
1
𝐴=−
6
5𝑥 − 15
𝑑𝑥
2
𝑥 − 𝑥 − 12
5𝑥 − 15
𝐴
𝐵
=
+
2
𝑥 − 𝑥 − 12
𝑥−4
𝑥+3
5𝑥 − 15 = 𝐴 𝑥 + 3 + 𝐵(𝑥 − 4)
Let 𝑥 = 4
5 4 − 15 = 𝐴(7)
5
𝐴=
7
Let 𝑥 = −3
5 −3 − 15 = 𝐵(−7)
30
𝐵=
7
5𝑥 − 15
𝑑𝑥 =
2
𝑥 − 𝑥 − 12
5/7
30/7
+
𝑑𝑥
𝑥−4
𝑥+3
𝟓
𝟑𝟎
𝐥𝐧 𝒙 − 𝟒 +
𝐥𝐧 𝒙 + 𝟑 + 𝑪
𝟕
𝟕
3𝑥 − 8
𝑑𝑥
2
𝑥 − 4𝑥 − 5
3𝑥 − 8
𝐴
𝐵
=
+
𝑥 2 − 4𝑥 − 5 𝑥 − 5 𝑥 + 1
3𝑥 − 8 = 𝐴 𝑥 + 1 + 𝐵 𝑥 − 5
Let x=-1
3 −1 − 8 = 𝐵 −6
∴ 𝐵=
11
6
3𝑥 − 8
𝑑𝑥 =
2
𝑥 − 4𝑥 − 5
Let x=5
3 5 −8=𝐴 6
7
11
6 + 6 𝑑𝑥
𝑥−5 𝑥+1
7
11
ln 𝑥 − 5 + ln 𝑥 + 1 + 𝐶
6
6
∴
𝐴=
7
6
𝑑𝑦
6𝑥 2 − 8𝑥 − 4
= 2
𝑑𝑥
𝑥 −4 𝑥−1
𝑥 2 − 4 𝑥 − 1 = (𝑥 + 2)(𝑥 − 2)(𝑥 − 1)
2
6𝑥 − 8𝑥 − 4
𝐴
𝐵
𝐶
=
+
+
𝑥2 − 4 𝑥 − 1
𝑥+2 𝑥−2 𝑥−1
𝟔𝒙𝟐 − 𝟖𝒙 − 𝟒 = 𝑨 𝒙 − 𝟐 𝒙 − 𝟏 + 𝑩 𝒙 + 𝟐 𝒙 − 𝟏 + 𝑪(𝒙 + 𝟐)(𝒙 − 𝟐)
𝟔𝒙𝟐 − 𝟖𝒙 − 𝟒 =
𝑨 𝒙 − 𝟐 𝒙 − 𝟏 + 𝑩 𝒙 + 𝟐 𝒙 − 𝟏 + 𝑪(𝒙 + 𝟐)(𝒙 − 𝟐)
Let 𝑥 = −2
36 = 12𝐴
6 −2
∴
Let 𝑥 = 2
∴
2
− 8 2 − 4 = 𝐵(4)(1)
𝐵=1
Let 𝑥 = 1
−6 = −3𝐶
− 8 −2 − 4 = 𝐴 (−4)(−3)
𝐴=3
6 2
4 = 4𝐵
2
6 1
∴
2
− 8 − 4 = 𝐶(3)(−1)
𝐶=2
6𝑥 2 − 8𝑥 − 4
3
1
2
=
+
+
2
𝑥 −4 𝑥−1
𝑥+2 𝑥−2 𝑥−1
6𝑥 2 − 8𝑥 − 4
=
2
𝑥 −4 𝑥−1
3
1
2
+
+
𝑥+2 𝑥−2 𝑥−1
3 ln 𝑥 + 2 + ln 𝑥 − 2 + 2 ln 𝑥 − 1 + 𝐶
End of Lesson
Partial Fraction Decomposition WS (1-11 odd)
In all of our examples thus far, the degree of the
numerator has been less than the degree of the
denominator.
If it is the case that the degree of the numerator is
greater than or equal to the degree of the
denominator, you must reduce using “polynomial”
long division.
The next few slides will help you to review this
technique…….
Long “Polynomial” Division Review
𝒙𝟐 − 𝟗𝒙 − 𝟏𝟎
𝒙+𝟏
𝒙 + 𝟏 𝒙𝟐 − 𝟗𝒙 − 𝟏𝟎
2
𝑥 + 9𝑥 + 14
𝑥+7
𝒙 + 𝟕 𝒙𝟐 + 𝟗𝒙 + 𝟏𝟒
3𝑥 3 − 5𝑥 2 + 10𝑥 − 3
3𝑥 + 1
𝟑𝒙 + 𝟏 𝟑𝒙𝟑 − 𝟓𝒙𝟐 + 𝟏𝟎𝒙 − 𝟑
3𝑥 3 − 5𝑥 2 + 10𝑥 − 3
dx
3𝑥 + 1
7
𝑥 − 2𝑥 + 4 −
𝑑𝑥
3𝑥 + 1
2
𝟏 𝟑
𝟕
𝟐
𝒙 − 𝒙 + 𝟒𝒙 − 𝐥𝐧 𝟑𝒙 + 𝟏 + 𝑪
𝟑
𝟑
2𝑥 3 − 9𝑥 2 + 15
2𝑥 − 5
Since no 𝒙 we must
Put 𝟎𝒙.
2𝑥 − 5 2𝑥 3 − 9𝑥 2 + 0𝑥 + 15
2𝑥 3 − 9𝑥 2 + 15 𝑑𝑥
2𝑥 − 5
𝑥2
1 3
𝑥
3
10
− 2𝑥 − 5 −
𝑑𝑥
2𝑥 − 5
− 𝑥 2 − 5𝑥 − 5 ln 2𝑥 − 5 + 𝐶
4
3
4𝑥 + 3𝑥 + 2𝑥 + 1
𝑥2 + 𝑥 + 2
Since no 𝒙𝟐 we must
Put 𝟎𝒙𝟐
𝑥 2 + 𝑥 + 2 4𝑥 4 + 3𝑥 3 + 0𝑥 2 + 2𝑥 + 1
End of Lesson
Homework:
Partial Fraction Decomp. Worksheet (14, 15)
Orange Book Section 6.5 P. 369
(5, 7, 8, 15-21 odd)
Steps to Integrating by PFD
1. If degree of numerator is greater than or equal
to degree of denominator, then use long division
to reduce.
2. Write out or setup the equation as a sum of
fractions with unknown numerators.
3. Solve for the unknown numerators.
4. Integrate the resulting equation.