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John Zhu

John Zhu

Integration By Partial Fractions

Slide Duration:

Table of Contents

Section 1: Prerequisites
Parametric Curves

10m 10s

Intro
0:00
Parametric Equations
0:23
Familiar Functions
0:32
Parametric Equation/ Function
0:56
Example 1: Graph Parametric Equation
1:48
Example 2
4:30
Example 3
6:01
Example 4
7:12
Example 5
8:10
Polar Coordinates

14m 54s

Intro
0:00
Polar Coordinates
0:09
Definition
0:11
Example
0:23
Converting Polar Coordinates
1:49
Example: Convert Polar Equation to Cartesian Coordinates
3:06
Example: Convert Polar Equation to Cartesian Coordinates
5:21
Example: Find the Polar Representation
6:51
Example: Find the Polar Representation
8:39
Example: Sketch the Graph
10:02
Vector Functions

12m 5s

Intro
0:00
Vector Functions
0:11
2 Parts
0:18
Example 1
1:44
Example 2
3:52
Example 3
4:59
Example 4
6:03
Example 5
8:00
Example 6
9:23
Example 7
10:16
Section 2: Differentiation
Parametric Differentiation

13m 15s

Intro
0:00
Parametric Differentiation
0:15
Example 1
1:16
Example 2
1:54
Example 3
3:15
Example 4
4:59
Parametric Differentiation: Position, Speed & Acceleration
7:45
Example 5
8:32
Example 6
10:37
Polar Differentiation

10m 14s

Intro
0:00
Polar Differentiation
0:11
Goal
0:13
Method
0:24
Example 1
0:52
Example 2
4:27
Example 3
7:03
L'Hopital's Rule

6m 43s

Intro
0:00
L'Hopitals Rule
0:09
Conditions
0:14
Limit
0:36
Example 1
1:21
Example 2
2:38
Example 3
4:09
Example 4
5:01
Section 3: Integration
Integration By Parts

19m 4s

Intro
0:00
Integration by Parts
0:12
Formula
0:31
How It's Derived From Product Rule
1:09
Integration by Parts: Rules to Follow
3:24
Recap of the Rules
4:13
Example 1
4:29
Example 2
7:04
Example 3
8:40
Example 4
13:48
Example 5
15:36
Integration By Partial Fractions

22m 4s

Intro
0:00
Integration by Partial Fractions: Goal
0:08
Example 1
2:10
Example 2
7:10
Example 3
10:06
Example 4
14:23
Example 5
16:26
Improper Integrals

19m 1s

Intro
0:00
Improper Integrals
0:06
3 Steps
1:11
Example 1
1:46
Example 2
3:20
Example 3
6:05
Example 4
9:02
Example 5
11:21
Example 6
15:54
Section 4: Applications of Integration
Logistic Growth

19m 16s

Intro
0:00
Logistic Growth Function
0:07
Defining Variables
0:40
Equation Parts
1:51
Logistic Growth
2:04
Example 1
2:59
Example 2
7:13
Example 3
11:29
Example 4
15:21
Arc Length for Parametric & Polar Curves

17m 40s

Intro
0:00
Arc Length
0:13
Arc Length of a Normal Function
0:24
Example
1:27
Example 2: Arc Length for Parametric Curves
3:31
Example 3: Arc Length for Parametric Curves
4:23
Example 4: Arc Length for Parametric Curves
8:05
Example 5: Arc Length for Parametric Curves
12:22
Example 6: Arc Length for Polar Curves
15:36
Example 7: Arc Length for Polar Curves
16:03
Area for Parametric & Polar Curves

4m 33s

Intro
0:00
Area for Parametric Curves: Parametric Function
0:10
Example 1: Area for Parametric Curves
0:35
Area for Parametric Curves: Polar Function
2:50
Example 2: Area for Polar Curves
3:18
Section 5: Sequences, Series, & Approximations
Definition & Convergence

7m 10s

Intro
0:00
Sequences: Definition
0:09
Definition
0:31
Example 1
1:07
Sequences: Convergence
2:02
Example 1
2:52
Example 2
3:36
Example 3
4:47
Example 4
6:16
Geometric Series

10m 59s

Intro
0:00
Geometric Series
0:07
Definition
0:24
Expanded Form
0:39
Convergence Rules
1:00
Example 1: Convergence
1:22
Example 2: Convergence
2:36
Example 3: Convergence
3:45
Sum of Series
5:04
Sum of First n Terms
5:14
Sum of Series
5:24
Example 1: Sum of Series
5:39
Example 2: Sum of Series
7:15
Example 3: Sum of Series
8:24
Example 4: Sum of Series
9:36
Harmonic & P Series

3m 58s

Intro
0:00
Harmonic Series
0:08
P-Series
1:17
Example 1: P-Series Test
2:22
Example 2: P-Series Test
3:06
Integral Test

11m 31s

Intro
0:00
Integral Test
0:09
Example 1
0:54
Example 2
4:00
Example 3
7:59
Example 4
9:47
Comparison Test

7m 47s

Intro
0:00
Comparison Test
0:10
Example 1
1:07
Example 2
2:33
Example 3
4:20
Example 4
6:29
Limit Comparison Test

9m 40s

Intro
0:00
Limit Comparison Test
0:09
Conditions
0:31
Example 1
1:01
Example 2
2:53
Example 3
4:15
Example 4
6:19
Ratio Test

11m 20s

Intro
0:00
Ratio Test
0:09
Example 1
0:57
Example 2
2:55
Example 3
6:27
Example 4
8:36
Alternating Series

9m 11s

Intro
0:00
Alternating Series
0:08
Convergence Test
0:59
Example 1
1:27
Example 2
2:42
Example 3
4:57
Example 4
6:37
Absolute Convergence

7m 13s

Intro
0:00
Absolute Convergence
0:12
Example 1
0:52
Example 2
3:42
Example 3
5:21
Power Series Convergence

12m 52s

Intro
0:00
Power Series
0:09
Definition
0:19
Radius & Interval of Convergence
2:07
Example 1
2:28
Example 2
4:18
Example 3
6:20
Example 4
9:11
Taylor Series

15m 11s

Intro
0:00
Taylor Series
0:06
MacLaurin Series
0:45
Example 1
1:02
Example 2
2:45
Example 3
6:04
Example 4
8:23
Example 5
11:49
Power Series Operations

16m 40s

Intro
0:00
Operations
0:07
Example 1: Substitution
1:05
Example 2: Substitution
3:41
Example 3: Differentiation/ Integration
5:39
Example 4: Differentiation/ Integration
9:55
Example 5: Multiplying/ Dividing
12:32
Example 6: Multiplying/ Dividing
14:50
Lagrange Error

7m 9s

Intro
0:00
Power Series: Lagrange Error
0:06
Lagrange Remainder
0:21
Lagrange Error Bound
0:50
Example 1
1:06
Example 2
3:27
Section 6: Practice Test
AP Calc BC Practice Test Section 1: Multi Choice Part 1

15m 45s

Intro
0:00
Practice Problem 1
0:10
Practice Problem 2
1:19
Practice Problem 3
2:33
Practice Problem 4
5:25
Practice Problem 5
6:32
Practice Problem 6
9:11
Practice Problem 7
11:31
Practice Problem 8
13:08
Practice Problem 9
14:16
AP Calc BC Practice Test Section 1: Multi Choice Part 2

21m 38s

Intro
0:00
Practice Problem 10
0:12
Practice Problem 11
3:03
Practice Problem 12
4:53
Practice Problem 13
6:56
Practice Problem 14
9:10
Practice Problem 15
12:17
Practice Problem 16
14:43
Practice Problem 17
16:18
Practice Problem 18
18:34
AP Calc BC Practice Test Section 1: Multi Choice Part 3

17m 31s

Intro
0:00
Practice Problem 19
0:09
Practice Problem 20
3:03
Practice Problem 21
4:07
Practice Problem 22
5:41
Practice Problem 23
8:48
Practice Problem 24
11:16
Practice Problem 25
12:53
Practice Problem 26
14:12
Practice Problem 27
15:21
AP Calc BC Practice Test Section 1: Multi Choice Part 4

14m 17s

Intro
0:00
Practice Problem 28
0:08
Practice Problem 29
1:19
Practice Problem 30
2:15
Practice Problem 31
4:19
Practice Problem 32
6:54
Practice Problem 33
8:13
Practice Problem 34
9:03
Practice Problem 35
10:21
Practice Problem 36
11:42
AP Calc BC Practice Test Section 1: Multi Choice Part 5

22m 45s

Intro
0:00
Practice Problem 37
0:10
Practice Problem 38
4:09
Practice Problem 39
7:21
Practice Problem 40
9:42
Practice Problem 41
11:35
Practice Problem 42
14:01
Practice Problem 43
15:50
Practice Problem 44
16:43
Practice Problem 45
19:35
AP Calc BC Practice Test Section 1: Free Response Part 1

9m 55s

Intro
0:00
Practice Problem 1a
0:08
Practice Problem 1b
1:30
Practice Problem 1c
3:08
Practice Problem 2a
3:50
Practice Problem 2b
5:40
Practice Problem 2c
7:24
AP Calc BC Practice Test Section 1: Free Response Part 2

10m 8s

Intro
0:00
Practice Problem 3a
0:09
Practice Problem 3b
3:01
Practice Problem 4a
3:55
Practice Problem 4b
5:43
AP Calc BC Practice Test Section 1: Free Response Part 3

13m 58s

Intro
0:00
Practice Problem 5a
0:07
Practice Problem 5b
2:21
Practice Problem 5c
4:21
Practice Problem 6a
6:16
Practice Problem 6b
8:57
Practice Problem 6c
11:03
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Lecture Comments (1)

0 answers

Post by Junhua Liu on June 5, 2018

Why is it wrong to put the 5 in 4/5 on the denominator with the (x+3) and (x-2)? When I put the 5 on the bottom, I got -4/5ln(5x+15)+4/5ln(5x-2).

Integration By Partial Fractions

  • Integrating by partial fractions:
    • Modify denominator into factored and multiple terms
    • Multiply constants A, B, C… into corresponding denominator terms
    • Rearrange terms to match original numerator
    • Set-up system of equations
    • Solve for all constants
    • Integrate using fundamental integral rules
  • Know the process very well, as this rarely changes for the AP exam

Integration By Partial Fractions

Expand [4/((x + 1)(x + 2))] using Partial Fractions
  • Setup form
  • [4/((x + 1)(x + 2))] = [A/(x + 1)] + [B/(x + 2)]
  • Multiply each side by the common demoninator
  • (x + 1)(x + 2)[4/((x + 1)(x + 2))] = (x + 1)(x + 2)( [A/(x + 1)] + [B/(x + 2)] )
  • 4 = A(x + 2) + B(x + 1)
  • Set x = − 1 to solve forA
  • 4 = A( − 1 + 2) + B( − 1 + 1)
  • 4 = A
  • Set x = − 2 to solve for B
  • 4 = A( − 2 + 2) + B( − 2 + 1)
  • 4 = − B
  • − 4 = B
Solve partial fractions
[4/((x + 1)(x + 2))] = [A/(x + 1)] + [B/(x + 2)] = [4/(x + 1)] − [4/(x + 2)]
Solve ∫ [4/((x + 1)(x + 2))] dx
  • Use partial fractions to integrate
  • ∫ [4/((x + 1)(x + 2))] dx = ∫ [4/(x + 1)] − [4/(x + 2)] dx
  • = ∫ [4/(x + 1)] dx − ∫ [4/(x + 2)] dx
  • Use Chain Rule to integrate
  • ∫ [4/(x + 1)] dx = 4ln|x + 1| + C
∫ [4/(x + 2)] dx = 4ln|x + 2| + C∫ [4/(x + 1)] dx − ∫ [4/(x + 2)] dx = 4ln|x + 1| − 4ln|x + 2| + C
∫0e [32/((x + 1)(x + 2))] dx =
  • Note the factors of 32
  • ∫0e [32/((x + 1)(x + 2))] dx = 8∫0e [4/((x + 1)(x + 2))] dx
  • = 8[ 4ln|x + 1| − 4ln|x + 2| ]0e
  • = 8(4ln1 − 4ln2 − 4ln(1 + e) + 4ln(2 + e))
= 32( − ln2 − 2ln( (1 + e)(2 + e) )
Expand [(x − 9)/(x2 − 10x + 21)] using Partial Fractions
  • Simplify the denominator
  • [(x − 9)/(x2 − 10x + 21)] = [(x − 9)/((x − 7)(x − 3))]
  • Setup form
  • [(x − 9)/((x − 7)(x − 3))] = [A/(x − 7)] + [B/(x − 3)]
  • Multiply each side by the common demoninator
  • (x − 7)(x − 3)[(x − 9)/((x − 7)(x − 3))] = (x − 7)(x − 3)( [A/(x − 7)] + [B/(x − 3)] )
  • x − 9 = A(x − 3) + B(x − 7)
  • Set x = 7 to solve for A
  • 7 − 9 = A(7 − 3) + B(7 − 7)
  • − 2 = 4A
  • − [1/2] = A
  • Set x = 3 to solve for B
  • 3 − 9 = A(3 − 3) + B(3 − 7)
  • − 6 = − 4B
  • [2/3] = B
Solve partion fractions
[(x − 9)/((x − 7)(x − 3))] = [A/(x − 7)] + [B/(x − 3)] = [( − [1/2])/(x − 7)] + [([2/3])/(x − 3)] = [2/3]( [1/(x − 3)] ) − [1/2]( [1/(x − 7)] )
Solve ∫ [(x − 9)/(x2 − 10x + 21)] dx
  • Use partial fractions to integrate
  • ∫ [(x − 9)/(x2 − 10x + 21)] dx = ∫ [2/3]( [1/(x − 3)] ) − [1/2]( [1/(x − 7)] ) dx
  • = ∫ [2/3]( [1/(x − 3)] ) dx − ∫ [1/2]( [1/(x − 7)] ) dx
  • = [2/3]∫ ( [1/(x − 3)] ) dx − [1/2]∫ ( [1/(x − 7)] ) dx
  • Use Chain Rule to integrate
  • [2/3]∫ ( [1/(x − 3)] ) dx = [2/3]ln|x − 3| + C
  • [1/2]∫ ( [1/(x − 7)] ) dx = [1/2]ln|x − 7| + C
[2/3]∫ ( [1/(x − 3)] ) dx − [1/2]∫ ( [1/(x − 7)] ) dx = [2/3]ln|x − 3| + C − [1/2]ln|x − 7| + C
Expand [2x/(x2 + 2x − 8)] using Partial Fractions
  • Simplify the denominator
  • [2x/(x2 + 2x − 8)] = [2x/((x + 4)(x − 2))]
  • Setup form
  • [2x/((x + 4)(x − 2))] = [A/(x + 4)] + [B/(x − 2)]
  • Multiply each side by the common demoninator
  • (x + 4)(x − 2)[2x/((x + 4)(x − 2))] = (x + 4)(x − 2)( [A/(x + 4)] + [B/(x − 2)] )
  • 2x = A(x − 2) + B(x + 4)
  • Set x = − 4 to solve for A
  • 2( − 4) = A( − 4 − 2) + B( − 4 + 4)
  • − 8 = − 6A
  • [4/3] = A
  • Set x = 2 to solve for B
  • 2(2) = A(2 − 2) + B(2 + 4)
  • 4 = 6B
  • [2/3] = B
  • Solve partial fractions
  • [2x/((x + 4)(x − 2))] = [A/(x + 4)] + [B/(x − 2)]
  • = [([4/3])/(x + 4)] + [([2/3])/(x − 2)]
= [4/3]( [1/(x + 4)] ) + [2/3]( [1/(x − 2)] )
Solve ∫ [2x/(x2 + 2x − 8)] dx
  • Use partial fractions to integrate
  • ∫ [2x/(x2 + 2x − 8)] dx = ∫ [4/3]( [1/(x + 4)] ) + [2/3]( [1/(x − 2)] ) dx
  • = ∫ [4/3]( [1/(x + 4)] ) dx + ∫ [2/3]( [1/(x − 2)] ) dx
  • = [4/3]∫ ( [1/(x + 4)] ) dx + [2/3]∫ ( [1/(x − 2)] ) dx
  • Use Chain Rule to integrate
  • [4/3]∫ ( [1/(x + 4)] ) dx = [4/3]ln|x + 3| + C
  • [2/3]∫ ( [1/(x − 2)] ) dx = [1/2]ln|x − 2| + C
[4/3]∫ ( [1/(x + 4)] ) dx + [2/3]∫ ( [1/(x − 2)] ) dx = [4/3]ln|x + 4| + [2/3]ln|x − 2| + C
Expand [1/(x2 − 3x)] using Partial Fractions
  • Simplify the denominator
  • [1/(x2 − 3x)] = [1/((x)(x − 3))]
  • Setup form
  • [1/((x)(x − 3))] = [A/x] + [B/(x − 3)]
  • Multiply each side by the common demoninator
  • (x)(x − 3)[1/((x)(x − 3))] = (x)(x − 3)( [A/x] + [B/(x − 3)] )
  • 1 = A(x) + B(x − 3)
  • Set x = 3 to solve for A
  • 1 = A(3) + B(3 − 3)
  • 1 = 3A
  • [1/3] = A
  • Set x = 0 to solve for B
  • 1 = A(0) + B(0 − 3)
  • 1 = − 3B
  • − [1/3] = B
[1/((x)(x − 3))] = [A/x] + [B/(x − 3)] = [([1/3])/x] + [( − [1/3])/(x − 3)] = [1/3]( [1/x] ) − [1/3]( [1/(x − 3)] )
Setup ∫ [lnx/((x − 3)2)]dx into Integration by Parts
  • Let u = lnx, dv = [1/((x − 3)2)]
  • du = [dx/x]
  • v = [( − 1)/(x − 3)] + C
Setup Integration by Parts
∫ [lnx/((x − 3)2)]dx = lnx( [( − 1)/(x − 3)] ) − ∫ [( − 1)/(x − 3)]( [dx/x] ) + C
Solve ∫ [lnx/((x − 3)2)] dx
  • Apply Integration by Parts
  • ∫ [lnx/((x − 3)2)] = lnx( [( − 1)/(x − 3)] ) − ∫ [( − 1)/(x − 3)]( [dx/x] ) + C
  • Apply Partial Fractions
  • lnx( [( − 1)/(x − 3)] ) − ∫ [( − 1)/(x − 3)]( [dx/x] ) + C
  • = lnx( [( − 1)/(x − 3)] ) + ∫ [dx/((x)(x − 3))] + C
  • = − [lnx/(x − 3)] + ∫ [1/3]( [1/x] ) − [1/3]( [1/(x − 3)] ) + C
  • = − [lnx/(x − 3)] + [1/3]∫ [1/x]dx − [1/3]∫ [1/(x − 3)]dx + C
= − [lnx/(x − 3)] + [1/3]ln|x| − [1/3]ln|x − 3| + C

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Integration By Partial Fractions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Integration by Partial Fractions: Goal 0:08
  • Example 1 2:10
  • Example 2 7:10
  • Example 3 10:06
  • Example 4 14:23
  • Example 5 16:26

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