For more information, please see full course syllabus of College Calculus: Level II

For more information, please see full course syllabus of College Calculus: Level II

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### 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__:Use

*u*= denominator first.Then separate into .

On the first part, use ln |

*u*|.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

- 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

[4/((x + 1)(x + 2))] = [A/(x + 1)] + [B/(x + 2)] = [4/(x + 1)] − [4/(x + 2)]

_{}

^{}[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

^{e}

_{0}[32/((x+1)(x+2))] dx

- f(x) = [32/((x+1)(x+2))]
- ∫f(x) dx = 32·(ln(x+1)−ln(x+2))

^{e}

_{0}f(x) dx = 32(−ln(e+2)+ln(e+1)+ln(2))

^{2}− 10x + 21)] using Partial Fractions

- Simplify the denominator
- [(x − 9)/(x
^{2}− 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)] =

_{}

^{}[(x − 9)/(x

^{2}− 10x + 21)] dx

- Use partial fractions to integrate
- ∫
_{}^{}[(x − 9)/(x^{2}− 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

_{}

^{}( [1/(x − 3)] ) dx − [1/2]∫

_{}

^{}( [1/(x − 7)] ) dx = [2/3]ln|x − 3| + C − [1/2]ln|x − 7| + C

^{2}+ 2x − 8)] using Partial Fractions

- Simplify the denominator
- [2x/(x
^{2}+ 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)]

_{}

^{}[2x/(x

^{2}+ 2x − 8)] dx

- Use partial fractions to integrate
- ∫
_{}^{}[2x/(x^{2}+ 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

_{}

^{}( [1/(x + 4)] ) dx + [2/3]∫

_{}

^{}( [1/(x − 2)] ) dx = [4/3]ln|x + 4| + [2/3]ln|x − 2| + C

^{2}− 3x)] using Partial Fractions

- Simplify the denominator
- [1/(x
^{2}− 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

_{}

^{}[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

∫

_{}

^{}[lnx/((x − 3)

^{2})]dx = lnx( [( − 1)/(x − 3)] ) − ∫

_{}

^{}[( − 1)/(x − 3)]( [dx/x] ) + C

_{}

^{}[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

*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

### 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
- Additional Example 4
- Additional Example 5

### College Calculus 2 Online Course

I. Advanced Integration Techniques | ||
---|---|---|

Integration by Parts | 24:52 | |

Integration of Trigonometric Functions | 25:30 | |

Trigonometric Substitutions | 30:09 | |

Partial Fractions | 41:22 | |

Integration Tables | 20:00 | |

Trapezoidal Rule, Midpoint Rule, Left/Right Endpoint Rule | 22:36 | |

Simpson's Rule | 21:08 | |

Improper Integration | 44:18 | |

II. Applications of Integrals, part 2 | ||

Arclength | 23:20 | |

Surface Area of Revolution | 28:53 | |

Hydrostatic Pressure | 24:37 | |

Center of Mass | 25:39 | |

III. Parametric Functions | ||

Parametric Curves | 22:26 | |

Polar Coordinates | 30:59 | |

IV. Sequences and Series | ||

Sequences | 31:13 | |

Series | 31:46 | |

Integral Test | 23:26 | |

Comparison Test | 22:44 | |

Alternating Series | 25:26 | |

Ratio Test and Root Test | 33:27 | |

Power Series | 38:36 | |

V. Taylor and Maclaurin Series | ||

Taylor Series and Maclaurin Series | 30:18 | |

Taylor Polynomial Applications | 50:50 |

### 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 x ^{2} + 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 x ^{2} + 4x + 5 down into linear terms.*0052

*What do we do instead?*0058

*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 = x ^{2} + 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/x ^{2} + 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 x ^{2} + 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 x ^{2} + 4x + 5.*0240

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

*So, 4/2 = 2, 2 ^{2} = 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 w^{2} + 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 x ^{2} + 4x + 5.*0370

*Then we will add on a constant as before.*0382

*There are a couple of important points about this example.*0391

*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 - u ^{2}.*0453

*You use secant when you have something like u ^{2} - 1.*0460

*Now, either one of those can be factored.*0465

*u ^{2} - 1 factors as (u-1)(u+1).*0469

*1- u ^{2} 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

*x ^{2} + 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, 8x ^{2} + 30x + 30/x^{3} + 6x + 13x + 10 = a/x + 2,*0276

*Then it would be nice if we could factor x ^{2} + 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

*x ^{2} + 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 × (x ^{2} + 4x + 5).*0339

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

*The x ^{3} term that we started with.*0348

*What you would get is a × x ^{2} + 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 8x ^{2} + 30x + 30.*0367

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

*So, my x ^{2}, I am going to have an a here,*0384

*Then from the b and the x here, we get a b x ^{2}.*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 8x ^{2} +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 x ^{2} term, an x term, and a constant term.*0455

*On both sides.*0465

*The x ^{2} 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/x ^{2} + 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/x ^{2} + 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 denominator*0621

*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(x ^{2} + 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 x ^{2} + 5x + 6.*0084

*We are going to forget about calculus for a while, and just write this as 1/(x ^{2} + 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/(x ^{2} + 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=1*0258

*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) dX*0307

*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

*(2x ^{2} - 7x)/(x^{2}+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 (x ^{2}-6x+8) into (2x^{2}-7x) and I am going to stick on 0 as a constant term there.*0480

*We look at x ^{2} and 2x^{2} 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 2x ^{2}-12x+16, subtract those the 2x^{2} cancel, subtracting is the same as changing signs and adding.*0505

*So, minus 7x + 12x gives us 5x - 16.*0518

*We can write (2x ^{2} - 7x)/(x^{2}-6x+8) as our answer 2 plus a remainder term (5x-16)/(x^{2}-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

*(x ^{2}-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) ^{2}*0914

*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 u ^{2}, 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

3 answers

Last reply by: Dr. William Murray

Mon Sep 5, 2016 9:46 AM

Post by Peter Ke on July 27, 2016

For example 3, just a curious question what happens if you square the x-3 under the A? So it will be A/(x-3)^2? Will you still get the same answer?

1 answer

Last reply by: Dr. William Murray

Thu Aug 13, 2015 8:51 PM

Post by Micheal Bingham on July 31, 2015

How do we know whether to use partial fractions or to complete the square? For the first question I used the method for completing the square and used a trigonometric substitution and came to a completely different answer.

3 answers

Last reply by: Dr. William Murray

Tue Feb 11, 2014 4:07 PM

Post by Angela Patrick on January 27, 2014

professor could you contact educator and correct some of the practice questions. On one the the answer is simplified incorrectly and on another they say 6=4b so b=2/3

1 answer

Last reply by: Dr. William Murray

Mon Oct 21, 2013 8:02 PM

Post by Ana Moga on October 20, 2013

Thanks for the great lectures ...you are an amazing Prof.

I have a quick naive request: do you mind clarifying how do you get (x+2)^2 +1 from x^2+4x+5 (example IV- minute 4:35) :) thank you very much for your time and help

1 answer

Last reply by: Dr. William Murray

Fri Jul 5, 2013 9:07 AM

Post by William Dawson on June 25, 2013

The algebra is definitely the hardest part, and it's always good to show it if it involves any significant rearrangement. There is nothing 'elementary' about algebra at this level either, whatever some _____ might say in the responses, in trying to be cute.

2 answers

Last reply by: Dr. William Murray

Mon May 13, 2013 10:42 AM

Post by Megan Kell on February 16, 2012

When he moved from (A+B)x = 5x and -4A-2B = -16 to A+B = 5 and -4A-2B = 16, he forgot the negative sign on the 16. Is this correct, or am I missing something?

1 answer

Last reply by: Dr. William Murray

Mon May 13, 2013 10:38 AM

Post by Kathy Hodge on November 7, 2011

For drill problems on Integration by Parts and other integration techniques, go to

http://archives.math.utk.edu/visual.calculus/4/index.html

However, they don't go into this topic in as much depth -- or as many types of denominators as Professor Murray.

1 answer

Last reply by: Dr. William Murray

Mon May 13, 2013 10:34 AM

Post by ahmed hassan on May 1, 2011

You can skip all the steps and give us the answer!!

1 answer

Last reply by: Dr. William Murray

Mon May 13, 2013 10:29 AM

Post by Daniel Brook on February 21, 2011

Thanks professor, your algebraic technique of sperating A and B was different from the way I learned in class and it was really helpful.

1 answer

Last reply by: Dr. William Murray

Mon May 13, 2013 10:24 AM

Post by James Reaper on February 27, 2010

In example 2 the negative sign wasn't carried through or dropped by dividing by -1.

2 answers

Last reply by: Dr. William Murray

Sun Sep 9, 2012 9:02 PM

Post by lynette stevenson on February 16, 2010

excellent professor, however he skipped steps he should explain the algerbra in more detail