Sign In | Subscribe
INSTRUCTORS Raffi Hovasapian John Zhu

Enter your Sign on user name and password.

Forgot password?
  • Follow us on:
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of AP Calculus AB
  • Discussion

  • Download Lecture Slides

  • Table of Contents

  • Transcription

Lecture Comments (2)

1 answer

Last reply by: Professor Hovasapian
Mon Jul 25, 2016 6:57 PM

Post by Peter Ke on July 22, 2016

Hello, I am confused about how you got the top portion of the equation, but I do understand how you got the bottom portion.


Please explain!

Integration by Partial Fractions I

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 I 0:11
    • Recall the Idea of Finding a Common Denominator
    • Decomposing a Rational Function to Its Partial Fractions
    • 2 Types of Rational Function: Improper & Proper
  • Improper Rational Function 7:26
    • Improper Rational Function
  • Proper Rational Function 11:16
    • Proper Rational Function & Partial Fractions
    • Linear Factors
    • Irreducible Quadratic Factors
  • Case 1: G(x) is a Product of Distinct Linear Factors 17:10
  • Example I: Integration by Partial Fractions 20:33
  • Case 2: D(x) is a Product of Linear Factors 40:58
  • Example II: Integration by Partial Fractions 44:41

Transcription: Integration by Partial Fractions I

Hello, welcome back to, and welcome back to AP Calculus.0000

Today, we are going to be talking about a technique called integration by partial fractions.0004

Let us jump right on in.0010

I think I will stick with black, for the time being.0013

Recall the concept of finding a common denominator.0017

Recall the idea of finding a common denominator.0024

Something like, if we had 2/ x + 2 - 3/ x – 3.0039

We want to find the common denominator.0048

The common denominator is going to be x + 2 × x – 3.0050

What we end up doing, we multiply this, so we get 2 × x – 3.0054

Nice, basic mathematics from long ago.0058

3 × x + 2/ x + 2 × x – 3.0061

Of course, we multiply everything out.0071

We end up with 2x - 6 - 3x - 6/ x² 2x - x – 6.0075

2x, we get - x - 12/ x² – x – 6.0093

This is the same as that, except under a common denominator.0103

Now what if we have to evaluate?0107

What if we were asked to evaluate the following integral.0112

The integral of – x - 12/ x² – x - 6 dx.0121

If we had a procedure from going backward, from this to this, we can write this integral this way,0131

and then just integrate both of these which we know how to do.0158

What if we had a procedure from going backward from - x - 12/ x² - x - 6 to 2/ x + 2 - 3/ x – 3.0162

Then, we would do the integral of 2/ x + 2 dx - the integral of 3/ x - 3 dx.0183

We know what this equal already.0199

We already know that this = 3 × natlog of x + 2 - 3 × natlog of x - 3 + c.0204

What we want to do is we want to find a way of taking a rational function,0221

working backward, breaking it up into what we call partial fractions.0226

We want to do a partial fraction decomposition and then,0230

we want to integrate each one of those partial fractions based on techniques that we have already developed,0234

either straight integration or any of the other techniques that we have developed.0238

That is what we want.0243

This procedure, if we are going backward, it is what we will develop in this lesson.0249

This procedure, if we are going backward is what we develop here.0255

Some of you may have seen it, some of you perhaps you have not seen it.0268

Not a problem, it is what we develop here.0271

We call it decomposing a rational function.0277

Rational just means you have a numerator which is a polynomial, the denominator is a polynomial.0287

Rational function into its partial fractions.0291

That is why we call it the partial fraction decomposition.0303

There are two types of rational functions.0310

An example of one is, let us say x³ + 5/ x² + 9.0331

Another would be x² + 9/ x³ + 5.0340

Here, this is called improper.0348

The reason it is improper is because the degree of the numerator.0353

The 3 is bigger than the degree of the denominator.0359

The degree of the numerator is bigger than the highest degree of the denominator, that is called improper.0362

This one, where the degree of the numerator is less than the degree of the denominator, that is called proper.0371

Proper because the degree of numerator is less than the degree of the denominator.0380

Any proper rational function can be decomposed into its partial fractions, any one.0393

If the rational function that we have to deal with is already proper, it is fine,0421

we just subject it to this procedure that we are going to develop to decompose it to its partial fractions.0426

How do we deal with it, if we have a rational function which is not proper?0432

We do a long division and we turn it into something proper.0436

The question is, but how do we handle an improper rational function?0447

We perform a long division, a polynomial long division, if you remember from algebra.0463

We rewrite this numerator which is a function of x/ denominator which is a function of x.0474

Something that is improper, we do the long division, and we get a quotient0484

which is a function of x that is the one on top + the remainder that we get,0489

which is the one in the bottom/ the divisor which is the denominator.0497

We rewrite it, we do a long division and express it this way.0505

We take the integral of this.0509

The integral of this, whatever we need to do here, this now is going to be proper.0511

We can subject that to a partial fraction decomposition.0516

Here, this is improper, this is just our quotient.0521

This is now going to be proper which we can then decompose.0532

Proper, this, we decompose.0548

Let us do our example.0558

Let us have a rational function, let us go back to black.0560

I have got a rational function f(x) is equal to x³ + 5/ x² + 5.0566

This is improper, let us do the long division.0580

The long division is x² + 5 x³ + 0 x², I have to fill in every space, + 0x + 5.0582

What times x² gives me x³?0601

It is going to be x.0603

x × x² is going to be x³.0604

x5, I’m going to put the 5x here.0608

I'm going to change the sign.0611

This cancels with that, I'm left with -5x + 5.0616

This is my quotient, this is my remainder.0622

This is my divisor, my denominator.0626

We have this thing, turns into is equal to x + -5x + 5/ x² + 5.0630

There you go, this is proper and I could subject it to a partial fraction decomposition which is the next thing we are going to do.0649

That is it, anytime you have an improper, do the long division.0660

You are going to get a quotient + a proper fraction.0663

Let us go ahead and talk about our different case now.0672

Let us say every proper rational function can be decomposed into its partial fraction sum.0701

In other words, it can be expressed as a sum of partial fractions.0727

Every proper one can, every proper fractions, where the denominators of the partial fractions,0741

of the individual partial fractions are linear factors and/or irreducible quadratic factors.0761

Let me explain what this means.0785

You actually know what this means already, you have seen it a thousand times.0788

Anytime you factor a polynomial, you can always factor it and get a bunch of factors multiplied by each other.0799

Those factors are either going to be linear, in other words,0811

the exponent of the x is going to be 1 or they are going to be quadratic.0814

The exponent is going to be 2.0819

Irreducible means you hit a quadratic function that you cannot factor any further.0821

It is always going to be that.0826

It is always going to be 1 or 2.0827

You can always break it down.0828

You are never going to have something that is going to be cubic, or quartet, or quintet.0829

It is always going to be a linear factor which is x¹ or a quadratic factor x².0833

Here is what they look like.0842

A linear factor looks like this, partial fractions of linear factors.0843

It is a fraction so it is going to have some numerator.0851

A partial fraction, let us read this again.0856

Every rational function can be decomposed into partial fractions sum, that can be expressed as a sum of partial fractions,0858

where the denominators of the partial fractions are linear factors or irreducible quadratics.0871

The linear factor is going to be some a which is a constant/ ax + b, to some m power.0875

Linear factor looks like this.0886

Linear means the power of x is 1, x + 6, 2x – 2, 3x + 15, these are linear factors.0888

Quadratic factor looks like this.0901

The exponent on the x itself, not this one up here.0920

These are multiplicities, that is however many times the root shows up, when you factor a polynomial.0922

A linear is when the exponent here is 1.0929

Quadratic is when the exponent on the x itself is 2, it is a quadratic factor.0936

Notice, this is a power of 1, the number on top has to be a constant.0940

It has to be 1° less.0953

A quadratic partial fraction is quadratic in the denominator, the x, it is linear in the numerator.0956

It is always going to be like that.0966

In your partial fractions, these things, however many there are.0969

1, 2, 3, 4, 5, the denominator is linear, the top is just a number.0973

If the denominator is quadratic, the top is some linear function.0978

It is going to be bx + c.0982

It is going to be 1° less.0984

The letters are just constants.0989

Do not worry, everything will make a lot of sense in just a minute.0999

It is actually very simple, it is quite algorithmic.1001

There is no problem.1004

We distinguish four different cases.1006

That is fine, I will go ahead and do it this way.1013

We distinguish four cases.1014

In case 1, we have a rational function where, in other words f(x)/ g(x).1025

Case 1 is when we factor the denominator which is what we are always going to do first, it takes a rational function.1044

You are going to break down the denominator as much as possible.1050

You are going to factor it out, as much as possible.1053

G(x) is a product, if when you factor it, you find that it is a product of distinct linear factors,1057

case 1, g(x) is a product of distinct linear factors.1079

In other words, this means this, thru words.1083

g(x), when it is factored is equal to some a1x + b1 × a2x + b2 × some a sub nx + b sub n.1092

In other words, no factors are repeated.1111

These a's and bs are different.1114

a1 and b1, a2 and b2, an, bn, they are all different.1115

In other words, it does not have a multiplicity of more than one, that is what this means.1121

Distinct linear factors, it means a multiplicity that factor occurs only once.1126

No factor is repeated.1132

In this case, the partial fraction decomp looks like this.1140

f(x)/d(x) is going to equal some a1/ a1x + b1 + a2/ a2x + b2 + so on and so forth, + a sub n/ a sub nx + b sub n.1161

Our task, in other words, we are going to write it out like this.1192

Our task is going to be find what a1 through an are.1197

Our task is to find a1, a2, and so on, all the way to an.1205

All we have done is factor the denominator into its factors.1216

Written those factors, we know that it breaks up into some composition of partial fractions.1219

We need to find what the numerators of those partial fractions are.1225

An example will make this clear.1229

Let us work in blue.1235

Find the integral of x² + 3x - 10/ 3x³ + 8x² - 3x.1237

The first thing we always do is factor the denominator, that is what we want.1243

We want to break it up into its factors.1247

They are going to be either linear or they are going to be quadratic.1249

We factor as far as we can.1252

We always start by factoring the denominator.1255

If it is already factored, you are done, that part is taken care for you.1266

All the factors will be written.1269

Sometimes it is given that way.1271

We always start by factoring the denominator, as far as possible.1272

Again, the factoring will always be linear, quadratic.1286

You might be only linear, you might be only quadratic.1290

You might combination linear and quadratic, but it is always going to be linear or quadratic.1292

This first case that we are dealing with is they are all linear and they are all distinct, multiplicity 1.1297

Let us take our denominator which is 3x³.1306

We have got 3x³ + 8x² - 3x.1309

This is equal to x × 3x² + 8x – 3, that is equal to x × 3x - 1 × x + 3.1320

There you go, that is our full factorization.1341

Linear, linear, linear, 3 factors.1345

Linear, exponent is 1, linear, exponent is 1.1349

They are all distinct.1354

This, this, and this are completely separate, they are completely distinct.1355

We have three linear factors.1359

Our partial fraction decomp looks like this.1368

It looks like, we have x² + 3x - 10/ 3x³ + 8x² - 3x is equal to, it is equal to some a,1378

some constant a/ the first factor x + some constant b/ the second factor 3x - 1 + some constant c/ the third factor x + 3.1401

That is what we meant.1416

This is a common denominator.1420

This, this, this, this and this are actually the same.1423

I know that there are three fractions, that when I combine them to get a common denominator, I'm going to get that.1428

I'm working backward.1434

My task is to find a, find b, find c.1436

There are three distinct linear factors.1440

I put each one of the denominator as a separate fraction, a, b, c.1442

I need to find a, b, c.1446

We want to find a, b, and c, that is what we want to do.1450

Let us go ahead and do that.1462

This is how we start, we write out the decomposition.1464

What we are going to do, this is equal to that.1472

I need to find this, this, this.1475

What I'm going to do is I'm going to express the right side, in terms of the common denominator.1478

The common denominator, I know what the common denominator is.1485

The common denominator is this × this × this.1488

That means I'm going to multiply a by this and this.1492

I’m going to multiply b by this and this.1495

I’m going to multiply c by this and this.1498

When I do that, you will see what happens when I do that.1502

Now we write the right side.1511

We write now the right side, in terms of a common denominator.1518

We have x² + 3x - 10/ 3x³ + 8x² - 3x is equal to a × 3x - 1 × x + 31534

+ b × x × x + 3 + c × x × 3x - 1/ x × 3x - 1 × x + 3.1563

This is equal to that.1588

Since this is equal to that, we can sort of ignore them, that means the numerator is equal to the numerator.1592

Let us expand the numerator and see what we get.1598

When I expand the numerator, this is going to be a ×,1602

that is fine, I will go ahead and just work with that.1622

Let us go ahead and expand that.1624

We have a × this is going to be 3x² + 8x - 3 + bx² + 3bx + 3cx² – cx.1626

It is going to be 3ax² + 8ax – 3a + bx² + 3bx + 3cx² – cx.1661

I’m going to combine terms, combine common terms.1680

In other words, anyone that has x² and x² in it.1693

Let me do this one in blue.1699

I lost my little color changing thing.1701

We are stuck with red for the rest of the time, not a problem.1705

Combine the common terms, 3ax² and x².1707

I have got x² × 3a, I’m going to pullout the coefficients.1720

3a takes care of that one and x² is + b.1729

I’m just combining common terms, by combining the coefficients 3a and b.1734

I’m just writing it with a coefficient on this side, instead of the other side.1740

I hope that is not too much of a problem.1743

I have another one for x².1748

I have got 3c, sorry about that.1750

ab + 3c +, I have x terms, x.1755

I have got 8a, it takes care of that one.1763

+ 3b – c, that takes care of the x term.1771

I have +, the only term I am left with is this -3a.1778

I have expanded the numerator, now I have this thing.1788

We have this, we now have left side which is x² + 3x - 10/ 3x³ + 8x² - 3x,1792

all of that is equal to x² × 3a + b + 3c + x × 8a1810

+ 3b - c + -3a/ x × 3x - 1 × x + 3.1827

We expressed it, multiplied it out, now I have this.1843

I will go back to red.1848

The denominators are now the same, except one is non factor form and one was factored form, but they are the same thing.1850

Because the denominators are the same thing, I can ignore them.1856

That means the numerators are the same thing.1859

That means this numerator is equal to that numerator.1864

Because they are equal, every term on the left has a corresponding term on the right.1867

Here we have an x² term, the coefficient is 1.1873

That means over here, there is an x² term, that means this is its coefficient, it is equal to 1.1877

This x term, its coefficient is 3.1884

This x term, that is this thing, this is equal to 3.1888

-10 is the number, -3a, this is equal to -10.1892

I set equal the coefficients of corresponding terms.1898

I get a system of three equations and three unknowns.1902

I’m going to solve for a, b, and c.1904

That is how I do this.1907

Because the numerators are equal, now what I have got is the following.1910

x² + 3x - 10 is equal to x² × 3a + b + 3c + x × 8a + 3b - c + -3a.1914

The only way that these two are equal, I know they are equal if the denominators are equal.1940

The only way left and right side are equal is if corresponding coefficients are equal.1954

In other words, 3a + b + 3c has to equal 1.1976

8a + 3b - c is equal to 3 - 3a is equal to -10.1989

-3a – 10, that implies that a = 10/3.2009

I have already found my a.2016

I’m going to put that a into this equation.2020

I’m going to put this a into this equation.2024

I'm going to get two equations and two unknowns.2026

I wonder if I should go through process.2032

That is fine, this is probably a good review.2033

I’m going to put a in here.2035

This is going to be 3 × 10/3 + b + 3c is equal to 1.2037

And then, I have 8 × 10/3 + 3b - c is equal to 3.2047

This implies, when I multiply, move things over, I'm going to end up with b + 3c is equal to -9.2059

I’m going to end up with 3b - c is going to equal -71/3.2071

I'm going to multiply the top by -3, that gives me -3b - 9c = 27.2088

I will leave the other one alone.2108

3b - c = -71/3.2109

Add them straight, I end up with -10c is going to end up equaling 10/3, that means c is going to equal -1/3.2116

I found my c, now I put that into one of these equations.2130

I will put it into this one, I get 3b, - and -1/3 = -71/3.2136

I get 3b + 1/3 = -71/3.2149

I get b is equal to 8/9, when I solve.2155

I found a, I found b, I found c.2166

Now I put them back into my partial fraction decomposition that I actually wrote first.2172

Let us go back.2177

What happened here, I did something here.2184

Now I have, remember the partial fraction composition that we,2198

Something is wrong with my, let us try black.2206

I have got x² + 3x - 10/, remember our original, 3x³ + 8x² - 3x.2213

We wrote the partial fraction decomposition.2225

We said that that is equal to a/x + b/ 3x - 1 + c/ x + 3, that was our partial fraction decomposition.2227

We have manipulated this partial fraction decomposition.2240

We set corresponding coefficients equal to each other.2242

We solve the system of equations.2245

We found a, b, and c.2246

We found that it is equal to 10/3 / x + 8/9 / 3x - 1 + -1/3.2249

We usually leave it like this.2262

With + in between, we leave the - on top, / x + 3.2263

This is our partial fraction decomposition of that.2269

Our original was, what we started out doing, we wanted to find the integral of this.2273

That is just the integral of this, we decomposed it.2282

It is the integral of that, 10/3 / x + 8/9 / 3x - 1 + -1/3 / x + 3 dx.2286

That is equal to 10/3 × the integral of 1/x dx + 8/9 × the integral of 1/ 3x - 1 dx -1/32306

× the integral of 1/ x + 3 dx this = 10/3 ×,2315

we did the partial fraction decomposition because now these are all logarithms.2335

That is why we did it.2339

The natlog of the absolute value of x + 8/9 × 1/3 × natlog of 3x – 1.2340

1/3 comes from the fact that this is 3x – 1, we use a u substitution real quickly.2357

u = 3x – 1, du = 3dx, dx = du/3.2364

Therefore, this integral is, the integral of 1/u du.2374

du, the 1/3 comes out, that gets pulled out as a 1/3.2379

This is -1/3 × the natlog of x + 3 + c.2383

There is your partial fraction decomposition.2392

This partial fraction decomposition are very long and they are very tedious.2394

That is just the nature of the game.2403

I would recommend using some online software, as far as solving the equations and unknowns2406

because you might have three equations and three unknowns.2411

You can work with it, sometimes, you may have 3, 4, 5.2414

I would definitely just use some software, in general, at least to get through the problems.2417

But you want to go through the partial fraction decomposition, to finding the common denominators,2422

setting corresponding things equal, that you want to do by hand.2427

That is it, that is partial fraction decomposition of a rational function,2432

where the denominator can be factored into three distinct linear factors.2439

There is no repeats, you do not have x².2445

You do not have 3x – 1³.2449

Multiplicity is 1 on each.2452

Let us talk about case 2.2455

Case 2 is when our rational function, this time, the denominator,2459

when we factor the denominator, it ends up being a product.2471

The factoring is a product of linear factors.2479

Again, we are sticking with linear factors, some repeated.2486

This time we have multiplicities which are going to be possibly greater than 1.2492

That is, the denominator is going to end up looking like, it is going to be some a1x + b1 raised to some power,2500

a2x + b2 raised to some power, a sub nx + bn raised to some power.2515

The partial fraction decomposition looks like this.2528

It looks like this, our nx/dx which is our rational function is equal to some constant,2543

a1/ the factor a1x + b1¹ + a2/ the same factor raised to the next higher power.2551

a1x + b1², and you keep going until you reach the nth power.2568

An nth power, that is just for the first factor.2591

And then, it is + b1/ the second, a2x + b2¹ + b2/ a 2x + b2²2594

+ a sub n a2x + b2 ⁺nth, + c1/ however many factors you have.2612

Each factor that you have, you are going to have that many terms all the way up to the nth power.2640

anx + bn, all of this will make sense when you see a problem, it is very simple, to the first power + c2/ a sub nx + b sub n.2646

Writing this out is exhausting.2659

+… + c ⁺p/ a sub nx + b sub n ⁺p.2662

Let us fluke with an example, I think that is the best way to make sense of this.2680

We have got ax + 14/ x + 5² 2x – 2x – 2.2685

Notice a couple of things, this one, the denominator is already factored for us.2693

We do not have to do the factoring.2696

The denominator is already factored, very nice.2701

You will notice that the factors are linear.2706

Exponent is 1, exponent is 1, they are linear.2709

One of the factors is repeated twice.2712

The x + 5 factor is repeated twice.2721

Therefore, the partial fraction decomposition for that factor is going to have two terms.2732

The partial fraction decomposition for this is going to have one term.2737

We are going to have a total of three terms.2740

Here is what it looks like.2742

It is going to be 8x + 14/ x + 5² × 2x - 2 is equal to, we will take the first factor.2744

We will deal with x + 5 first.2757

It is going to be a/ x + 5¹ + b/ x + 5².2759

Because that is 2 and that is 2, I can stop.2769

Now I move to the next factor, + c/ 2x – 2¹.2772

That is it, it is the first power that only shows up once.2779

Now we do what we do.2784

The least common denominator, I already know what that is.2792

The least common denominator is x + 5² × 2x - 2 which actually is that thing.2797

The numerator on the right becomes a × x + 5, because here x + 5 is only once.2816

The least common denominator has it twice.2832

a × x + 5 + b × 2x - 2 + c × x + 5² = ax + 5a.2837

I’m sorry, I forgot one.2859

It is a × x + 5 × 2x – 2.2867

I forgot one, + b × 2x - 2 + c × x + 5².2872

There we go, that equals a × 2x²,2883

You can see how things can go soft very quickly.2891

2x² + 8x - 10 + 2bx - 2b + c × x² + 10x + 252895

= 2ax² + 8ax - 10a + 2bx - 2b + cx² + 10cx + 25c.2916

Let us go ahead and combine.2942

x², x², we are going to take the x term, x term, x term.2944

And then, number, number, number.2954

Let us go ahead and do that.2959

Our final is going to be 8x + 14/ x + 5² × 2x – 2, that is going to equal x² × 2a + c + x ×,2962

when I combine those terms, 8a + 2b + 10c + -10a - 2b + 25c/ x + 5² × 2x – 2.2984

Because the denominators are the same, the numerators are the same,3009

and the only way they can be the same is if coefficients of corresponding terms are equal.3012

Therefore, 2a + c = over here there is no x² term which means it is 0.3021

8a + 2b + 10c, that is the coefficient of the x term.3035

The coefficient of the x term is equal to 8 - 10a - 2b + 25c is equal to 14.3042

There you go, this is the system of equations that you have to solve.3056

Now I'm not going to go through the process of solving, I hope you will forgive me.3060

I would like you to corroborate if you can, either by using software or doing it by hand.3063

I have done it here but I really do not want to write everything else.3067

I end up with a is equal to -11/36.3073

I get b is equal to 167/72.3080

I get c is equal to 11/18.3095

Those are my three coefficients.3100

Therefore, let us rewrite, we had 8x + 14 was our original rational function.3103

We have x + 5² × 2x – 2.3121

We said that it equals a/ x + 5 + b/ x + 5² + c/ 2x - 23127

that = -11/36 / x + 5 + 167/72 / x + 5² + 11/18 / 2x – 2.3140

The integral of this is the integral of this.3166

The integral of that breaks up into three integrals.3172

Let me go back to blue, = -11/36 × the integral of 1/ x + 5 dx + 167/72 ×3175

the integral of 1/ x + 5² dx + 11/18 × the integral of 1/ 2x - 2 dx.3201

The answers I get are – 11/36 × the natlog of x + 5, an absolute value, + 167/72 × 1/ -1 × 1/ x + 5.3215

I hope this integration does not throw you guys off.3238

+ 11/18, this one is just set u equal to x + 5, do a u substitution.3245

It ends up being the integral of u⁻² du, and then, some factor.3257

11/18 × ½, the natlog of 2x – 2.3264

This ½ term comes from the fact that that is a 2 + c.3270

There you go, very nice.3274

Very tedious but, beautiful process.3278

We have taken care of two cases here.3283

We have taken care of the case where the denominator factors into linear factors that are distinct.3285

We also did the linear factors that possibly some are repeated.3291

We are going to stop here.3295

In the next lesson, we are going to do where it factors into a linear and quadratic factors.3296

And then, linear and quadratic factors with some of the quadratic factors are now repeated.3302

With that, thank you so much for joining us here at

We will see you next time, bye.3310