Math

1. Sect. 7.4 Integration of Rational Functions by Partial Fractions
Review:
2
𝑥 − 2
+
3
𝑥 + 3
=
𝟐
(𝒙 − 𝟐)
∙
(𝒙 + 𝟑)
(𝒙 + 𝟑)
+
𝟑
(𝒙 + 𝟑)
∙
(𝒙 − 𝟐)
(𝒙 − 𝟐)
=
𝟐𝒙 + 𝟔 + 𝟑𝒙 − 𝟔
(𝒙 − 𝟐)(𝒙 + 𝟑)
=
𝟓𝒙
𝒙𝟐 + 𝒙 − 𝟔
Consider:
𝟓𝒙
𝒙𝟐 + 𝒙 − 𝟔
𝒅𝒙 =
𝟐
𝒙 − 𝟐
+
𝟑
𝒙 + 𝟑
𝒅𝒙
Partial Fractions
Easy to integrate
=
𝟐
𝒙 − 𝟐
𝒅𝒙 +
𝟑
𝒙 + 𝟑
𝒅𝒙

2. Steps
1. ___________________________________________________________________
2. ___________________________________________________________________
3. ___________________________________________________________________
4. ___________________________________________________________________
Divide, if improper. Degree of top > degree of bottom.
Factor denominator completely into irreducible factors.
Linear Factors. 𝒙 − 𝒄 ⟹ make a fraction
𝑨
𝒙 − 𝒄
I𝐟 𝐫𝐞𝐩𝐞𝐚𝐭𝐞𝐝 𝐥𝐢𝐧𝐞𝐚𝐫 𝐟𝐚𝐜𝐭𝐨𝐫𝐬 𝒙 − 𝒄 𝒏, make n fractions in the form …
𝑨
𝒙 − 𝒄 𝒏
+ ⋯ +
𝒁
(𝒙 − 𝒄)
+
𝑩
𝒙 − 𝒄 𝒏−𝟏 +
𝑪
𝒙 − 𝒄 𝒏−𝟐
Quadratic Factors. 𝒙𝟐 − 𝒄 ⟹ make a fraction
𝑨𝒙 + 𝑩
𝒙𝟐 − 𝒄
I𝐟 𝐫𝐞𝐩𝐞𝐚𝐭𝐞𝐝 𝐪𝐮𝐚𝐝𝐫𝐚𝐭𝐢𝐜 𝐟𝐚𝐜𝐭𝐨𝐫𝐬 𝒙𝟐
− 𝒄 𝒏, make n fractions in the form …
𝑨𝒙 + 𝑩
𝒙𝟐 − 𝒄 𝒏 + ⋯ +
𝒀𝒙 + 𝒁
(𝒙𝟐 − 𝒄)
+
𝑪𝒙 + 𝑫
𝒙𝟐 − 𝒄 𝒏−𝟏 +
𝑬𝒙 + 𝑭
𝒙𝟐 − 𝒄 𝒏−𝟐
Partial Fraction Decomposition
Decomposing a rational function into simpler functions to which basic integration formulas can
be applied.

3. Example 1:
5𝑥
𝑥2+𝑥−6
𝑑𝑥 1.
2.
3.
Is it improper? No
Factor denominator
𝑥2 + 𝑥 − 6 = 𝑥 + 3 𝑥 − 2
Partial fractions for
each linear factor
5𝑥
𝑥2 + 𝑥 − 6
=
𝐴
𝑥 + 3
+
𝐵
𝑥 − 2
Multiply by LCD to
both sides
𝟓𝒙 = 𝑨 𝒙 − 𝟐 + 𝑩 𝒙 + 𝟑
Substitute in
values for x to
solve for A & B
Let x = 2
𝟓 𝟐 = 𝑨 𝟐 − 𝟐 + 𝑩 𝟐 + 𝟑
𝟏𝟎 = 𝟓𝑩 𝟐 = 𝑩
Let x = - 3
𝟓 −𝟑 = 𝑨 −𝟑 − 𝟐 + 𝑩 −𝟑 + 𝟑
−𝟏𝟓 = −𝟓𝑨 𝟑 = 𝑨
So,
5𝑥
𝑥2 + 𝑥 − 6
𝑑𝑥 =
3
𝑥 + 3
𝑑𝑥 +
2
𝑥 − 2
𝑑𝑥
5𝑥
𝑥2 + 𝑥 − 6
=
𝟑
𝑥 + 3
+
𝟐
𝑥 − 2
and,
= 𝟑 𝐥𝐧 𝒙 + 𝟑 + 𝟐 𝐥𝐧 𝒙 − 𝟐 + 𝑪

4. Example 2:
4𝑥2
𝑥3+𝑥2−𝑥−1
𝑑𝑥 𝑥3 + 𝑥2 − 𝑥 − 1
Factor by grouping
= 𝑥2
𝑥 + 1 − 1 𝑥 + 1
= 𝑥 + 1 𝑥2 − 1
= 𝑥 + 1 𝑥 + 1 𝑥 − 1
= 𝑥 + 1 2 𝑥 − 1
4𝑥2
𝑥3 + 𝑥2 − 𝑥 − 1
=
𝐴
𝑥 − 1
+
𝐵
𝑥 + 1
+
𝐶
𝑥 + 1 2
𝟒𝒙𝟐
= 𝑨 𝒙 + 𝟏 𝟐 + 𝑩 𝒙𝟐
− 𝟏 + 𝑪 𝒙 − 𝟏
𝟒𝒙𝟐
= 𝑨𝒙𝟐
+ 𝟐𝑨𝒙 + 𝑨 + 𝑩𝒙𝟐
− 𝑩 + 𝑪𝒙 − 𝑪
Let x = 1
𝟒 𝟏 𝟐 = 𝑨 𝟏 + 𝟏 𝟐 + 𝑩 𝟏 𝟐 − 𝟏 + 𝑪(𝟏 − 𝟏)
𝟒 = 𝟒𝑨 𝟏 = 𝑨
𝟎 = 𝟏 − 𝑩 + 𝟐 𝑩 = 𝟑
Another solving technique.
Let x = – 1
𝟒 – 1 𝟐 = 𝑨 – 1 + 𝟏 𝟐 + 𝑩 – 1 𝟐 − 𝟏 + 𝑪(– 1 − 𝟏)
𝟒 = −𝟐𝑪 −𝟐 = 𝑪
Let x = 0
𝟒 0 𝟐 = 𝟏 0 + 𝟏 𝟐 + 𝑩 𝟎 𝟐 − 𝟏 + −𝟐(0 − 𝟏)
𝑨 = 𝟏 𝑪 = −𝟐
4𝑥2
𝑥3 + 𝑥2 − 𝑥 − 1
=
𝟏
𝑥 − 1
+
𝟑
𝑥 + 1
+
−𝟐
𝑥 + 1 2
Collect
x2 terms: 𝟒 = 𝑨 + 𝑩
x terms: 𝟎 = 𝟐𝑨 +𝑪
Constant
terms: 𝟎 = 𝑨 − 𝑩 − 𝑪
= 𝟒
= 𝟎
= 𝟎
rref
1 1 0
2 0 1
1 -1 -1
4
0
0
é
ë
ê
ê
ê
ù
û
ú
ú
ú
= 𝑨
= 𝑩
= 𝑪

5. Example 2:
4𝑥2
𝑥3+𝑥2−𝑥−1
𝑑𝑥
4𝑥2
𝑥3 + 𝑥2 − 𝑥 − 1
=
𝟏
𝑥 − 1
+
𝟑
𝑥 + 1
+
−𝟐
𝑥 + 1 2
4𝑥2
𝑥3 + 𝑥2 − 𝑥 − 1
𝑑𝑥 =
𝟏
𝑥 − 1
𝑑𝑥 +
𝟑
𝑥 + 1
𝑑𝑥 +
−2
𝑥 + 1 2 𝑑𝑥
= 𝐥𝐧 𝒙 − 𝟏 + 𝟑 ∙ 𝐥𝐧 𝒙 + 𝟏 − 𝟐
𝒙 + 𝟏 −𝟏
−𝟏
+ 𝑪
= 𝐥𝐧 𝒙 − 𝟏 + 𝟑 ∙ 𝐥𝐧 𝒙 + 𝟏 +
𝟐
𝒙 + 𝟏
+ 𝑪

6. Proper and Denominator is factored
𝑨𝒙 + 𝑩
𝒙𝟐 + 𝟗 𝟐 +
𝑪𝒙 + 𝑫
𝒙𝟐 + 𝟗
𝑥2
− 𝑥 + 9
𝑥2 + 9 2
=
𝑥2
− 𝑥 + 9 = 𝑨𝒙 + 𝑩 + 𝑪𝒙 + 𝑫 𝒙𝟐
+ 𝟗 = 𝑨𝒙 + 𝑩 + 𝑪𝒙𝟑
+ 𝑫𝒙𝟐
+ 𝟗𝑪𝒙 + 𝟗𝑫
x2 terms: 𝟏 = 𝑫
x terms: −𝟏 = 𝑨 + 𝟗𝑪
c terms: 𝟗 = 𝟗𝑫 + 𝑩
Collect
x3 terms: 𝟎 = 𝑪
𝑩 = 𝟎
𝟗 = 𝟗 + 𝑩
𝑥2
− 𝑥 + 9
𝑥2 + 9 2 𝑑𝑥 =
−𝟏𝑥
𝑥2 + 9 2 𝑑𝑥 +
𝟏
𝑥2 + 9
𝑑𝑥
u = _________
du = ___________
𝟐𝒙 𝒅𝒙
x2 + 9 = −
𝟏
𝟐
𝟏
𝒖𝟐
𝒅𝒖
= −
𝟏
𝟐
𝒖−𝟐𝒅𝒖
= −
𝟏
𝟐
∙
𝒖−𝟏
−𝟏 =
𝟏
𝟐 𝒙𝟐 + 𝟗
+
𝟏
𝟑
𝐭𝐚𝐧−𝟏
𝒙
𝟑
+ 𝑪
𝟏 𝟏
Integral Property
Example 3:
𝑥2−𝑥+9
𝑥2+9 2 𝑑𝑥
SHORT CUT!
𝒙𝟐
+ 𝟗
𝒙𝟐 + 𝟗 𝟐
+
−𝟏𝒙
𝒙𝟐 + 𝟗 𝟐
𝑥2
− 𝑥 + 9
𝑥2 + 9 2
= =
𝟏
𝒙𝟐 + 𝟗
+
−𝟏𝒙
𝒙𝟐 + 𝟗 𝟐

7. Example 4:
𝑥3+𝑥2+𝑥+2
𝑥2+1 𝑥2+2
𝑑𝑥 Proper and Denominator is factored
𝑥2 + 2
𝑥2 + 1 𝑥2 + 2
+
𝑥3 + 𝑥
𝑥2 + 1 𝑥2 + 2
𝑥3
+ 𝑥2
+ 𝑥 + 2
𝑥2 + 1 𝑥2 + 2
=
𝑥3
+ 𝑥2
+ 𝑥 + 2
𝑥2 + 1 𝑥2 + 2
𝑑𝑥 =
𝟏
𝑥2 + 1
𝑑𝑥 +
𝟏𝑥
𝑥2 + 2
𝑑𝑥
Integral Property
= 𝐭𝐚𝐧−𝟏 𝒙
u = _________
du = ___________
𝟐𝒙 𝒅𝒙
x2 + 2
=
𝟏
𝟐
𝟏
𝒖
𝒅𝒖 =
𝟏
𝟐
𝐥𝐧 𝒖
+
𝟏
𝟐
𝐥𝐧 𝒙𝟐 + 𝟐 + 𝑪
Can you see the short cut?
𝑥 𝑥2 + 1
1

8. Example 5:
2𝑥5−𝑥3−1
𝑥3−4𝑥
𝑑𝑥 NOT Proper! Need to Divide the polynomials.
x3
- 4x 2x5
- x3
-1
2x2
2x5
-8x3
+ 7
7x3
-1
28x -1
x x + 2
( ) x - 2
( )
+
7x3
- 28x
28x -1
2𝑥2 + 7 +
28𝑥 − 1
𝑥3 − 4𝑥
𝑑𝑥
= 2 𝑥2𝑑𝑥 + 7 𝑑𝑥 +
28𝑥 − 1
𝑥 𝑥 + 2 𝑥 − 2
𝑑𝑥
28𝑥 − 1
𝑥 𝑥 + 2 𝑥 − 2
=
𝑨
𝒙
+
𝑩
𝒙 + 𝟐
+
𝑪
𝒙 − 𝟐
28𝑥 − 1 = 𝑨 𝒙𝟐
− 𝟒 + 𝑩𝒙 𝒙 − 𝟐 + 𝑪𝒙 𝒙 + 𝟐
28𝑥 − 1 = 𝑨𝒙𝟐 − 𝟒𝑨 + 𝑩𝒙𝟐 − 𝟐𝑩𝒙 + 𝑪𝒙𝟐 + 𝟐𝑪𝒙 rref
1 1 1
0 -2 2
-4 0 0
0
28
-1
é
ë
ê
ê
ê
ù
û
ú
ú
ú
𝑨 + 𝑩 + 𝑪 = 𝟎
−𝟐𝑩 + 𝟐𝑪 = 𝟐𝟖
= −𝟏
−𝟒𝑨
= 2 𝑥2
𝑑𝑥 + 7 𝑑𝑥 +
𝟏
𝟒
𝒙
𝑑𝑥 +
−𝟓𝟕
𝟖
𝒙 + 𝟐
𝑑𝑥 +
𝟓𝟓
𝟖
𝒙 − 𝟐
𝑑𝑥

9. = 2 𝑥2
𝑑𝑥 + 7 𝑑𝑥 +
𝟏
𝟒
𝒙
𝑑𝑥 +
−𝟓𝟕
𝟖
𝒙 + 𝟐
𝑑𝑥 +
𝟓𝟓
𝟖
𝒙 − 𝟐
𝑑𝑥
=
𝟐𝒙𝟑
𝟑
−
𝟓𝟕
𝟖
𝐥𝐧 𝒙 + 𝟐
+ 𝟕𝒙 +
𝟏
𝟒
𝐥𝐧 𝒙 + 𝑪
+
𝟓𝟓
𝟖
𝐥𝐧 𝒙 − 𝟐