For more information, please see full course syllabus of AP Calculus AB

For more information, please see full course syllabus of AP Calculus AB

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### 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
- Integration by Partial Fractions I
- 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
- Proper Rational Function
- Case 1: G(x) is a Product of Distinct Linear Factors
- Example I: Integration by Partial Fractions
- Case 2: D(x) is a Product of Linear Factors
- Example II: Integration by Partial Fractions

- 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

### AP Calculus AB Online Prep Course

### Transcription: Integration by Partial Fractions I

*Hello, welcome back to www.educator.com, 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 quotient*0484

*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 + 3*1534

*+ 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 × 8a*1810

*+ 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 -1/3*2306

*× 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 unknowns*2406

*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 + b2²*2594

*+ 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 + 25*2895

*= 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 - 2*3127

*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 www.educator.com.*3308

*We will see you next time, bye.*3310

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.

"A(3X-1)(X+3)..... http://prntscr.com/bwa3pw

Please explain!