Professor Murray

Partial Fractions

Slide Duration:

Integration by Parts

24m 52s

Intro
0:00
Important Equation
0:07
Where It Comes From (Product Rule)
0:20
Why Use It?
0:35
Lecture Example 1
1:24
Lecture Example 2
3:30
Shortcut: Tabular Integration
7:34
Example
7:52
Lecture Example 3
10:00
Mnemonic: LIATE
14:44
Ln, Inverse, Algebra, Trigonometry, e
15:38
-1
-2
Integration of Trigonometric Functions

25m 30s

Intro
0:00
Important Equation
0:07
Powers (Odd and Even)
0:19
What To Do
1:03
Lecture Example 1
1:37
Lecture Example 2
3:12
Half-Angle Formulas
6:16
Both Powers Even
6:31
Lecture Example 3
7:06
Lecture Example 4
10:59
-1
-2
Trigonometric Substitutions

30m 9s

Intro
0:00
Important Equations
0:06
How They Work
0:35
Example
1:45
Remember: du and dx
2:50
Lecture Example 1
3:43
Lecture Example 2
10:01
Lecture Example 3
12:04
-1
-2
Partial Fractions

41m 22s

Intro
0:00
Overview
0:07
Why Use It?
0:18
Lecture Example 1
1:21
Lecture Example 2
6:52
Lecture Example 3
13:28
-1
-2
Integration Tables

20m

Intro
0:00
Using Tables
0:09
Match Exactly
0:32
Lecture Example 1
1:16
Lecture Example 2
5:28
Lecture Example 3
8:51
-1
-2
Trapezoidal Rule, Midpoint Rule, Left/Right Endpoint Rule

22m 36s

Intro
0:00
Trapezoidal Rule
0:13
Graphical Representation
0:20
How They Work
1:08
Formula
1:47
Why a Trapezoid?
2:53
Lecture Example 1
5:10
Midpoint Rule
8:23
Why Midpoints?
8:56
Formula
9:37
Lecture Example 2
11:22
Left/Right Endpoint Rule
13:54
Left Endpoint
14:08
Right Endpoint
14:39
Lecture Example 3
15:32
-1
-2
Simpson's Rule

21m 8s

Intro
0:00
Important Equation
0:03
Estimating Area
0:28
Difference from Previous Methods
0:50
General Principle
1:09
Lecture Example 1
3:49
Lecture Example 2
6:32
Lecture Example 3
9:07
-1
-2
Improper Integration

44m 18s

Intro
0:00
Horizontal and Vertical Asymptotes
0:04
Example: Horizontal
0:16
Formal Notation
0:37
Example: Vertical
1:58
Formal Notation
2:29
Lecture Example 1
5:01
Lecture Example 2
7:41
Lecture Example 3
11:32
Lecture Example 4
15:49
Formulas to Remember
18:26
Improper Integrals
18:36
Lecture Example 5
21:34
Lecture Example 6 (Hidden Discontinuities)
26:51
-1
-2
Section 2: Applications of Integrals, part 2
Arclength

23m 20s

Intro
0:00
Important Equation
0:04
Why It Works
0:49
Common Mistake
1:21
Lecture Example 1
2:14
Lecture Example 2
6:26
Lecture Example 3
10:49
-1
-2
Surface Area of Revolution

28m 53s

Intro
0:00
Important Equation
0:05
Surface Area
0:38
Relation to Arclength
1:11
Lecture Example 1
1:46
Lecture Example 2
4:29
Lecture Example 3
9:34
-1
-2
Hydrostatic Pressure

24m 37s

Intro
0:00
Important Equation
0:09
Main Idea
0:12
Different Forces
0:45
Weight Density Constant
1:10
Variables (Depth and Width)
2:21
Lecture Example 1
3:28
-1
-2
Center of Mass

25m 39s

Intro
0:00
Important Equation
0:07
Main Idea
0:25
Centroid
1:00
Area
1:28
Lecture Example 1
1:44
Lecture Example 2
6:13
Lecture Example 3
10:04
-1
-2
Section 3: Parametric Functions
Parametric Curves

22m 26s

Intro
0:00
Important Equations
0:05
Slope of Tangent Line
0:30
Arc length
1:03
Lecture Example 1
1:40
Lecture Example 2
4:23
Lecture Example 3
8:38
-1
-2
Polar Coordinates

30m 59s

Intro
0:00
Important Equations
0:05
Polar Coordinates in Calculus
0:42
Area
0:58
Arc length
1:41
Lecture Example 1
2:14
Lecture Example 2
4:12
Lecture Example 3
10:06
-1
-2
Section 4: Sequences and Series
Sequences

31m 13s

Intro
0:00
Definition and Theorem
0:05
Monotonically Increasing
0:25
Monotonically Decreasing
0:40
Monotonic
0:48
Bounded
1:00
Theorem
1:11
Lecture Example 1
1:31
Lecture Example 2
11:06
Lecture Example 3
14:03
-1
-2
Series

31m 46s

Intro
0:00
Important Definitions
0:05
Sigma Notation
0:13
Sequence of Partial Sums
0:30
Converging to a Limit
1:49
Diverging to Infinite
2:20
Geometric Series
2:40
Common Ratio
2:47
Sum of a Geometric Series
3:09
Test for Divergence
5:11
Not for Convergence
6:06
Lecture Example 1
8:32
Lecture Example 2
10:25
Lecture Example 3
16:26
-1
-2
Integral Test

23m 26s

Intro
0:00
Important Theorem and Definition
0:05
Three Conditions
0:25
Converging and Diverging
0:51
P-Series
1:11
Lecture Example 1
2:19
Lecture Example 2
5:08
Lecture Example 3
6:38
-1
-2
Comparison Test

22m 44s

Intro
0:00
Important Tests
0:01
Comparison Test
0:22
Limit Comparison Test
1:05
Lecture Example 1
1:44
Lecture Example 2
3:52
Lecture Example 3
6:01
Lecture Example 4
10:04
-1
-2
Alternating Series

25m 26s

Intro
0:00
Main Theorems
0:05
Alternation Series Test (Leibniz)
0:11
How It Works
0:26
Two Conditions
0:46
Never Use for Divergence
1:12
Estimates of Sums
1:50
Lecture Example 1
3:19
Lecture Example 2
4:46
Lecture Example 3
6:28
-1
-2
Ratio Test and Root Test

33m 27s

Intro
0:00
Theorems and Definitions
0:06
Two Common Questions
0:17
Absolutely Convergent
0:45
Conditionally Convergent
1:18
Divergent
1:51
Missing Case
2:02
Ratio Test
3:07
Root Test
4:45
Lecture Example 1
5:46
Lecture Example 2
9:23
Lecture Example 3
13:13
-1
-2
Power Series

38m 36s

Intro
0:00
Main Definitions and Pattern
0:07
What Is The Point
0:22
0:45
Interval of Convergence
2:42
Lecture Example 1
3:24
Lecture Example 2
10:55
Lecture Example 3
14:44
-1
-2
Section 5: Taylor and Maclaurin Series
Taylor Series and Maclaurin Series

30m 18s

Intro
0:00
Taylor and Maclaurin Series
0:08
Taylor Series
0:12
Maclaurin Series
0:59
Taylor Polynomial
1:20
Lecture Example 1
2:35
Lecture Example 2
6:51
Lecture Example 3
11:38
Lecture Example 4
17:29
-1
-2
Taylor Polynomial Applications

50m 50s

Intro
0:00
Main Formulas
0:06
Alternating Series Error Bound
0:28
Taylor's Remainder Theorem
1:18
Lecture Example 1
3:09
Lecture Example 2
9:08
Lecture Example 3
17:35
-1
-2
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• Related Books

Partial Fractions

Main formula:Something like

can be separated into

Hints and tips:

• You can always factor the denominator down into linear (degree one) and quadratic (degree two) factors. If you have factors that are cubic or higher, you should be able to factor more.

• To factor a cubic, test factors of the constant term divided by factors of the leading coefficient as roots. A quick way to check if they work is synthetic substitution (also known as synthetic division).

• Sometimes you can figure out the A and B by plugging in a particular value of x to both sides of the equation. (Choose a value that makes one of the expressions equal to 0.)

• If the degree of the numerator is the same or greater than the degree of the denominator, use long division first. Based on the results of your long division, split your integral into a polynomial part (easy to integrate) and a rational function that you can apply partial fractions to.

• If you have a quadratic in the denominator that doesn’t factor, then follow the following steps in order:

1. Use u = denominator first.

2. Then separate into .

3. On the first part, use ln |u|.

4. On the second, complete the square and use a trigonometric substitution. It should always be a tangent substitution. (If it is a sine or secant substitution, then you could have factored the quadratic.)

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
Solve ∫e0[32/((x+1)(x+2))] dx
• f(x) = [32/((x+1)(x+2))]
• ∫f(x) dx = 32·(ln(x+1)−ln(x+2))
e0f(x) dx = 32(−ln(e+2)+ln(e+1)+ln(2))
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
• [3/2] = B
• Solve Partial Fractions:
[(x − 9)/((x − 7)(x − 3))] =
[A/(x − 7)] + [B/(x − 3)] =
[( − [1/2])/(x − 7)] + [([3/2])/(x − 3)] =
[3/2]( [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.

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
• Overview 0:07
• Why Use It?
• Lecture Example 1 1:21
• Lecture Example 2 6:52
• Lecture Example 3 13:28

Transcription: Partial Fractions

OK, we are going to try a couple more examples of the partial fractions technique.0000

Remember that it is an algebraic technique of trying to separate a rational function,0005

Meaning a polynomial over a polynomial,0013

Into different pieces that are easier to integrate.0016

Remember that the very first step of partial fractions is to try to factor the denominator.0020

We are trying to factor x2 + 4x + 5,0027

Right away you get a problem.0035

It does not work.0038

So, even though this looks like a partial fractions problem,0044

It turns out that we are stuck at the quadratic level.0048

We cannot factor x2 + 4x + 5 down into linear terms.0052

Well, there is a two-step procedure for solving problems like this.0060

It is very important you do the steps in order.0063

The two steps are a u substitution, and then a trig substitution.0068

It is very important that you do those in order.0080

If you do not do those in the right order, you will end up giving yourself a lot of extra work.0083

Let us see how that plays out.0090

The first step here is let u = the denominator.0094

u = x2 + 4x + 5.0098

Then du = 2x + 4 dx.0104

So, what we would like to do is see if we have du in the numerator.0114

We do not, because we have 6x + 10, instead of 2x + 4.0120

So, we are going to write 6x + 10/x2 + 4x + 5.0127

We would really like to have du in the numerator.0136

I would really like to see du in the numerator, so I am going to write du in the numerator, 2x + 4.0145

That does not work, because we have a 6 over here.0152

To make that correct, I am going to put a 3 here.0156

That makes the 6's match.0160

We have 3 × 2x.0163

Unfortunately, that gives us 3 × 4 over here which is 12, which does not match the 10.0167

To make that match, I am going to subtract off a 2 x2 + 4x + 5.0175

So, 12 - 2 is 10, so it does match.0185

Now, we have 3 × 2x + 4 - 2.0192

We are going to attack those integral separately.0197

The left hand one we can use our u and du.0200

That turns into the integral of 3du/u.0205

Which is 3 × ln(u), and then we can substitute back,0212

And, we will be done with that one.0217

The right hand one is a little harder.0220

The right hand one is where we are going to use step 2.0222

We are going to use a trig substitution, but in order to do that, we have to complete the square on the denominator.0232

We are going to write the denominator x2 + 4x + 5.0240

Remember, completing the square, you take the middle term divide it by 2, and square it.0252

So, 4/2 = 2, 22 = 4, so I will write + 4.0262

And to make that accurate I have to write + 1 to match the 5 right there.0268

So we can write this as x + 2 2 + 1.0277

What I am going to do is a little substitution, u -, actually I do not want to use u because I already used u for the other integral.0283

So I will use w.0290

w = x + 2, and then dw = dx.0294

We have for the second integral, - the integral of 2 dw/x + what turned out to be (x + 2)2, so that is w2 + 1.0304

Then, the way you solve this is by letting w = tan(θ).0320

We learned that on the section on trig substitution and we actually did this integral a couple times,0327

So, I am not going to show you all the detail again.0331

You might remember all the details to the pattern, but if you do not,0334

Go ahead and do the substitution, put in dw, and you will get the answer quickly.0337

So, - 2 arctan(w), which in turn converts back to - 2 arctan(x + 2).0344

That was just the answer to the second integral.0363

We still have the first integral to combine with that.0367

That is 3 × ln of, u was x2 + 4x + 5.0370

Then we will add on a constant as before.0382

Remember, in all partial fractions problems, you try to factor the denominator first.0396

In this case, the denominator did not factor.0400

What we do is this 2 step procedure, it must be done in order.0406

First we do a u substitution where u = the denominator.0411

Then we do a trig substitution.0417

A secret trick here, is that when you do a trig substitution on one of these, it is always going to be a tangent substitution.0425

Remember we had three types of trig substitutions, tangent, sine, and secant.0432

But, when you have a problem like this, you will never get sine and you will never get secant.0440

Here is why you will never get sine and you will never get secant.0447

The reason you will never get sine or secant is because you use sine when you have something like 1 - u2.0453

You use secant when you have something like u2 - 1.0460

Now, either one of those can be factored.0465

u2 - 1 factors as (u-1)(u+1).0469

1- u2 factors as (1-u)(1+u).0474

So if you had either one of these trig substitutions,0481

It means you actually had something that would factor.0484

We said back here that the factoring does not work.0489

Essentially if you get over to the trig substitution step,0491

And you get a sine or a secant substitution, it means you screwed up somewhere.0496

So go back to the factoring step and see if you can factor it.0503

Check your work back there, it probably means you made a mistake.0506

If you did not make a mistake through any of your work, then you are guaranteed to get a tangent substitution at the end.0514

So you keep going through the tangent substitution and you end up with your answer.0522

We have got one more example on partial fractions.0000

It looks like the nastiest one of all.0004

It is a little bit more work than some of the others.0008

Remember, the first step is to check the degrees of the numerator and denominator.0010

Here we have got degree 2 in the numerator, and degree 3 in the denominator.0013

Remember what we are checking for.0020

We are checking to see if the numerator is bigger than or equal to the denominator.0024

Well, 2 is not bigger than 3, so it is OK, we do not have to go through long division.0027

So there is no long division which is a relief because these are pretty nasty polynomials.0033

So, we do not have to go through long division, but we do have to through and factor the denominator.0041

That is going to be a little bit of work, because this is a cubic.0050

Remember that the way you factor a cubic is you look at factors of the constant term, the 10.0058

Divided by factors of the leading term.0078

Which in this case is 1.0085

You look at + or - those factors and those are your possible candidates.0089

We have + or - factors of 10 could be 1, 2, 5, or 10.0099

Divided by factors of 1 is just + or - 1, so I will not even include that.0107

So we have 8 candidates for factors here.0113

I will show you how to test possible candidates.0118

Suppose we want x = 2.0122

Is that a possible candidate?0124

Well there is a trick you might have learned in high school algebra.0126

Where this is the same as checking whether x -2 is a factor of the polynomial.0133

What you do is you look at 2 and then you write down the coefficients, 1, 6, 13, and 10.0139

I am getting those from the coefficients of this polynomial.0149

Then you bring the one down, you multiply by 2, 2.0154

You add, so 6 + 2 = 8.0160

Multiply by 2, so that is 16, add, so that is 29.0164

Multiply by 2, 58, and then we add, and we get at the end 68.0170

So, that means that x = 2 is not a root there.0179

Or x-2 is not a factor.0186

Let us try something else, let us try x = -2.0189

Which means we are testing x+2 as a factor.0192

That means we put -2 in, we go through the same process, 1, 6, 13, and 10.0198

We bring the 1 down, multiply by -2, we add, so that is 4, 6 + -2 is 4.0206

Multiply by -2 = 8. Add and we get 5.0218

Multiply by -2, we get 10, and look, we got 0 at the end.0224

That tells us that x = -2 is a root and x+2 is a factor.0230

Moreover, it tells us that our polynomial factors into x+2 times, now you use these coefficients,0239

To get the other factor.0258

x2 + 4x + 5.0260

So this is all a technique that you learn in high school algebra,0265

But by the time that you get to Calculus 2 sometimes it is a little bit rusty.0271

What we are going to try to do is write, 8x2 + 30x + 30/x3 + 6x + 13x + 10 = a/x + 2,0276

Then it would be nice if we could factor x2 + 4x + 5 down further.0298

But notice that that does not factor anymore.0304

We are stuck with that as a quadratic term.0310

I will put that as my next denominator.0314

x2 + 4x + 5.0316

Because this is quadratic, I cannot just put b, I have to put bx + C.0320

Now, I am going to solve for a, b, and c.0330

If you combine these two terms over a common denominator,0333

The common denominator would be x + 2 × (x2 + 4x + 5).0339

It is the same as the denominator on the left.0342

The x3 term that we started with.0348

What you would get is a × x2 + 4x + 5.0352

+ bx + c × x + 2.0357

Because you would be multiplying these terms by x + 2.0364

All of that is equal to 8x2 + 30x + 30.0367

So, I am going to combine these 2 polynomials and separate them by power of x.0376

So, my x2, I am going to have an a here,0384

Then from the b and the x here, we get a b x2.0389

+, now let me see what we get for an x term, we get a 4a over here from the 4.0395

For an x term we get 2b + C.0404

The constant term is 5a + 2b, sorry not 2b.0415

The constant term is 2c.0431

That is still equal to 8x2 +30x + 30.0434

I am going to carry these equations over onto the next slide and we will see what we can do with them.0446

From the previous slide, we had an x2 term, an x term, and a constant term.0455

On both sides.0465

The x2 term on one side, it was a + b.0468

On the other side it was 8.0477

The x term was 4a + 2b + C = 30.0480

The constant term was 5a on one side, + 2C, and on the other side, = 30.0490

Here you have 3 equations and 3 unknowns.0500

It is a little messy but it is nothing impossible from high school algebra.0505

You could start with a quick substitution of a = 8 - b.0512

Plus that into the other two equations and then you get 2 equations in b and c.0521

Then you get a high school algebra where you are solving for 2 equations and 2 unknowns.0535

It is a little messy but not too bad.0539

I am going to skip over the details of that and just tell you what the answers come out to be.0542

You can try working it out on your own.0547

It turns out that a = 2, b = 6, and c = 10.0550

What that means is that our integral, remember we had a/x+2, so that is 2/x+2 + 6x + 10/x2 + 4x + 5.0557

We have got to integrate all of that.0584

What we get there, this is an integral we have seen before,0588

This is just 2 ln(x+2) abs(x+2),0595

6x + 10/x2 + 4x + 5.0600

That sounds more complicated, but we saw the technique to do that in the previous example.0604

Remember, there was a two-step technique for that, there was a u substitution,0611

U = the denominator0621

Then the second step, after you plough through the u substitution, was a trig substitution.0625

Remember that the trig substitution should always come out to be a tangent substitution,0633

Because if it is a sine or a secant substitution, it means you really could have factored the denominator.0639

We said here that this is a denominator that we cannot factor.0647

I rigged up the numbers so that it comes out exactly the same as the previous problem.0656

I am not going to go through solving this again, because this is the exact same as the example we did on the previous problem.0665

Let me show you just what the answer came out to be.0672

This came out to be 3 ln(x2 + 4x + 5).0675

That was the part that came out of the u substitution.0684

- 2 × arctan(x + 2).0689

You can look this up on the previous problem or you can work it out.0697

That is the answer that we got.0700

All of this put together is the answer for the whole integral.0703

Let us recap what we had here.0712

We had the integral of a polynomial over a cubic.0715

What you try to do with problems like these is you try to factor the cubic.0724

There are sort of two ways it could factor.0733

It could factor into three linear terms,0736

If so, then you try to split it up into a over 1 linear term + b over the next.0740

+ C over the next.0750

That did not happen for this problem.0754

This problem factored into x over a quadratic term.0757

The quadratic term did not factor.0762

What we had to do there was write it as a over the linear term,0767

+, we had to go with bx + C over the quadratic term.0775

Solve for a, b, and c.0783

Then the a part, integrating that is easy.0788

The b part is where we go through this 2 step procedure of a u substitution and a trig substitution.0789

That is the end of the lecture on partial fractions.0797

Thank you for watching educator.com.0799

Hi, this is educator.com, and we are going to learn today about partial fractions.0000

Partial fractions is a technique used to solve integrals but it is really grounded in algebra.0004

We are going to be doing a lot of algebra and not that much integration.0012

The point of partial fractions is to solve integrals where you have something like one polynomial, or maybe just a constant, divided by another polynomial.0018

What you are going to do is factor the denominator polynomial.0037

I have an example of a really horrific one here.0043

In practice, the ones you get on your homework, you will not have one this bad.0047

But this is kind of an example of the worst it could possibly be.0052

The denominator is a really complicated polynomial, but we factored it down.0055

You will factor the denominator polynomial, and then you will do some algebra to separate this fraction into other pieces, into simpler fractions that are supposed to be easier to integrate.0058

As usual, it is much easier to do this using examples rather than to simply talk about it in the abstract.0078

Let us go ahead and try some examples.0079

The numerator polynomial is just 1, the denominator is x2 + 5x + 6.0084

We are going to forget about calculus for a while, and just write this as 1/(x2 + 5x + 6)0092

Just doing some algebra, the urge here is to factor the denominator.0104

That is an easy one to factor, that factors into (x + 2)(x + 3).0108

What partial fractions does, is it tries to separate that into a constant divided by (x + 2) plus another constant divided by (x + 3).0116

We are going to try to figure out what those constants should be.0136

If you imagine putting these two terms on the right back together, we would combine that back over a common denominator, (x + 2) × (x + 3).0141

To put them over a common denominator we would have to write a × (x + 3) + b × (x + 2).0159

Remember this is still supposed to be equal to 1/(x2 + 5x + 6).0164

Effectively the numerator, a × (x+3) + b × (x+2) has to be equal to the numerator 1.0174

I am going to expand out the left hand side as a polynomial, that is ax + bx, so that is (a+b)x + 3a + 2b = 1.0189

I want to think of 1 as being a polynomial, so I want to think of that as 0x + 1.0203

Remember that whenever you have two polynomials equal to each other, that means their coefficients have to be equal.0212

That is saying the coefficient of x has to be equal on both sides, so (a+b)=0 and the constant coefficients have to be equal as well.0220

(3a+2b)=1.0230

This is now a linear system that you learned how to solve in high school algebra.0240

There are a number of different ways of doing this.0246

You can do linear combinations, you can do substitution, I think the easiest way of doing this is probably substitution.0256

If we write a=-b, and then plug that into the second equation, we get -3b+2b=1, so -b=1 so b=-1 and a=10258

That is just solving a system of variables and two unknowns just like you learned how to do in algebra 1 or 2 in high school.0280

So we take those numbers and we plug them back into our separation and we get that our integral is equal to two separate integrals.0295

a=1/(x+2) dX + b=-1/(x+3) dX0307

Those are both easy integrals to solve.0317

We can solve the first one by doing u=x+2 dU=dX, so we get the integral of dU/u, and that is equal to ln(abs(u)) which is the ln(abs((x+2))).0319

The second one is solved in exactly the same technique, so I will skip over the details of that.0342

This minus is from the negative sign above the ln(x+3) + a constant.0353

The integration there at the end was very easy.0364

The important part that we are trying to learn here is the algebra, where you start with a polynomial in the denominator, you factor it out and then you try to split this expression up into two parts with an a and a b.0370

Then, you combine those two parts back together, that gives you two equations for a and b.0386

You solve those two equations for a and b, you substitute them back into the original expression, and you get an integral that hopefully turns out to be easy to solve.0397

Let us try another example that gets a little more complicated.0407

Example 2 here, again we have a polynomial over a polynomial.0411

(2x2 - 7x)/(x2+6x+8)0415

There is a problem with this one, it will not work as well as the one on the previous page.0423

The problem is the degree of the two polynomials.0426

If the degree of the numerator is bigger than or equal to the degree of the denominator, then you can not go to partial fractions directly.0436

You have to do something else from high school algebra, which is long division of polynomials.0460

Let me show you how to do a long division here.0464

We are going to do (x2-6x+8) into (2x2-7x) and I am going to stick on 0 as a constant term there.0480

We look at x2 and 2x2 and we divide those into each other and we get 2, and then just like long division of numbers we multiply 2 by the whole thing here.0495

So, we get 2x2-12x+16, subtract those the 2x2 cancel, subtracting is the same as changing signs and adding.0505

So, minus 7x + 12x gives us 5x - 16.0518

We can write (2x2 - 7x)/(x2-6x+8) as our answer 2 plus a remainder term (5x-16)/(x2-6x+8).0527

Now, the important thing here is that the degree of the numerator is now 1, the degree of the denominator is now 2.0551

The denominator now has a bigger degree and partial fractions is going to work.0566

Remember, partial fractions does not work if the degree of the numerator is bigger than the degree of the denominator.0569

That is no longer true after we do polynomial division and we can expect the partial fractions to work.0580

I am not going to worry about the 2 for a while, because I know that 2 is going to be easy to integrate.0584

I am going to do partial fractions now on 5x-16 over that denominator, and I can factor that into (x-2) × (x-4).0598

I am going to try to separate that out into a/(x-2) + b/(x-4).0614

Just like we did before, we are going to end up with a × (x-4) + b × (x-2), that comes from trying to combine these terms over a common denominator, is equal to the numerator on the other side (5x-16).0626

(a+b) × (x=5x and -4a-2b=-16,0647

so a+b = 5 and -4a -2b = 16.0664

Again, you get a high school algebra problem, two equations and two unknowns, and you can solve that using any technique that you prefer from algebra.0678

I am just going to tell you the answer now, it turns out that a=3 and b=2.0687

You can check that in the two equations to see that it works.0700

Our integral now becomes the integral of 2, plus a = 3 so that is 3/(x-2), and b was 2 so that is 2/(x-4).0702

All that stuff that we did so far is just algebra, we still have to integrate this problem.0724

However, it is a very easy integral now.0731

The integral of 2 is just 2x.0733

We saw that is easy to integrate 1/(x + or - a constant), you just get ln(abs(x-2)) + 2(ln(abs(x-4)) + a constant.0740

Let us reiterate, the key thing that we learned from this example is that before you start doing partial fractions, you have to check the degree of the numerator and denominator.0761

If they are equal, or if the denominator is smaller, then you have got to do this step of long division or the partial fractions is not going to work for you.0782

After you do the long division, you get a part that is easy to integrate and a remainder term and the remainder term should work out for you when you try to do partial fractions.0796

The next example looks pretty similar to the previous examples but it has a little twist as you will see in a moment.0811

Again we start out by checking the degrees.0822

The degree is 1 in the numerator, the degree of the denominator, the power of the highest term, is 2.0824

So that is ok, the numerator has a smaller degree than the denominator so we expect partial fractions to work.0831

We are going to go ahead and try to factor the denominator.0843

(x2-6x+9), that factors as (x-3)2 and if you are going to try to separate that out using partial fractions, your first instinct might be to try a/(x-3) + b/(x-3) the same way we did the previous examples.0845

That does not work.0873

Here is why that does not work, Suppose you did that.0875

If you try to combine those back together you get (a+b) × (x-3) and that does not match, that cannot possibly match (3x-7)/(x-3)2.0878

If you naively try a/(x-3) + b/(x-3), the partial fractions is not going to work.0897

That would give you a big problem.0908

The secret to fixing this is to have a/(x-3) but b/(x-3)20914

That is kind of the secret to this problem, is to put in that square on (x-3).0931

Then if you try to find, if you try to match up the numerators, we get (3x-7)=a × (x-3).0935

To get the common denominator, and then + b because if you wanted to combine those two, our common denominator would be (x-3)2 and b is already over that denominator, so we would not have to multiply b by anything to get that common denominator.0948

Then, we get ax-3a + b = (3x-7), so ax plus, let me separate out terms (3a + b)=3x-7 and again you get two equations and two unknowns except this one is very easy, you get a=3 and minus 3a+b=-7.0970

Again, solve that using anything you want from high school algebra and of course you get a=3 and you plug that in, it works out to b=2.1003

So, our integral turns into the integral of a/(x-3) + , now b=2/(x-3)2, all of that times dX.1021

Now, we have learned already how to integrate 3/(x-3), that is ln(abs(x-3)).1045

2/(x-3)2, we have not seen that recently.1054

The way you do that is you make a little substitution, u = x-3, so dU = dX.1058

That gives you the integral of 2dU over u2, and you think of that as being u-2 and the integral of that is 2u-1/-1.1070

This gives you 3ln(abs(x-3)), I am going to pull this -1 outside so we get -2.1091

Now, u-1 = (x-3) and u-1 = 2/(x-3) and then I will tag on a constant.1102

OK, so the key observation in that way is that if you have a denominator that factors into a perfect square, you can not separate it out as you did in the previous examples.1125

You have to separate it out into (x-3) and an (x-3)2, and you still have an a and a b on both parts.1142

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