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 II

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
- Case 3: D(x) Contains Irreducible Factors 0:09
- Example I: Integration by Partial Fractions 5:19
- Example II: Integration by Partial Fractions 16:22
- Case 4: D(x) has Repeated Irreducible Quadratic Factors 27:30
- Example III: Integration by Partial Fractions 30:19

### AP Calculus AB Online Prep Course

### Transcription: Integration by Partial Fractions II

*Hello, welcome back to www.educator.com, and welcome back to AP Calculus.*0000

*Today, we are going to continue our discussion of integration by partial fractions.*0004

*Let us jump right on in.*0008

*If you remember in the last lesson, we did the first two of the four cases,*0011

*where once we factored the denominator of the rational expression as much as we can,*0016

*if we have all linear factors or if some of those linear factors are repeated.*0023

*Those are the first two cases that we dealt with last time.*0028

*Now in this lesson, we are going to talk about quadratic factors and repeated quadratic factors, the last two cases.*0031

*Let us start with case 3, I think I’m going to go ahead and work in blue.*0039

*Case 3 is when our rational function, some m(x)/d(x),*0044

*Again, this is just numerator and denominator as functions of x.*0051

*That is when the d(x) contains irreducible quadratic factors.*0055

*After you reduce, after you factored it as much as you can, one of the factors is quadratic.*0064

*In other words, the highest degree on the x, the highest exponent is a 2,*0068

*contains irreducible factors none of which is repeated.*0074

*The first one is going to be none repeating irreducible quadratic factors is repeated.*0090

*Let us recall irreducible means we cannot factor it any further.*0103

*We cannot factor the quadratic any further into linear factors.*0115

*That is our hope, we want to be able to factor something, any polynomial into, as many linear factors as possible.*0134

*As it turns out, ultimately, you can only go to linear and quadratic.*0141

*You can always you do that, but sometimes you cannot always factor the quadratic.*0145

*That is the irreducible.*0149

*Here is an example of irreducible is x² + 4.*0151

*This one, you cannot reduce any further.*0161

*Another example might be something like x² + 3x + 1.*0164

*You cannot factor it any further.*0172

*We said in the previous lesson that each linear factor ax + b gives a partial fraction of a/ ax + b.*0177

*If one of the factors is linear on top, the variable, the thing that we are looking for is just a constant, it is just a.*0212

*Now for quadratic factors, each factor of the form ax² + bx + c, because the quadratic factor is going to be some variation of this.*0219

*Sometimes the bx term would not be there.*0248

*Sometimes the c would not be there.*0250

*It is going to look something like that.*0255

*It gives ax + b/ ax² + bx + c.*0257

*On the top, our unknown, where it is actually going to be a full linear factor.*0267

*The thing to notice here is the denominators of degree 1, the numerator is 1° less.*0273

*That is why it is just a, it is x⁰.*0278

*Here the denominator, the highest degree is 2.*0281

*Therefore, on the numerator, it is going to be 1° less which means this type of function ax + b, a linear factor.*0285

*For linear factors, we put a on top, a constant.*0295

*For quadratic factors, we put ax + b.*0298

*Sorry, we use capitals, ax + b.*0301

*Where a, b, and c, and d, it could be, if you have another quadratic factor, it would be cx + d, ex + f, and so on.*0304

*Let us go ahead and do an example, I think it will make sense.*0315

*We want to evaluate the integral 12/ x - 2 × x² + 9.*0321

*The denominator here, again, the first thing we do is factor the denominator as much as we can.*0326

*Here the denominators are already factored.*0331

*x - 2 is our linear factor, x² + 9 is our irreducible quadratic factor.*0333

*This does not factor anymore.*0338

*If it were x² – 9, that is fine, we can do x + 3x - 3 but were stuck like this.*0339

*The denominators are already factored.*0346

*Now 12, this 12/ x - 2 × x² + 9, it is going to equal, we have a linear factor x – 2.*0358

*The partial fraction decomposition is going to be a/ x - 2 + this is our quadratic factor.*0373

*This is going to be x² + 9 and we put bx + c.*0381

*Our task is to find the a, find the b, and find the c, so that we have a partial fraction decomposition of our original rational function.*0386

*Once we separate that, we are going to integrate each one separately.*0395

*Let us go ahead and do that.*0398

*This is going to be, I actually solve for the least common denominator here.*0407

*This is going to be a × x² + 9 + bx + c × x - 2/ the least common denominator which is x - 2 × x² + 9.*0411

*We are going to concern ourselves only with the numerator.*0433

*Because now we have this equal to this, the denominators, this and this are the same,*0435

*that means that the numerator and the numerator are the same.*0447

*I'm just going to work with the numerators.*0451

*Once you actually do that on the right, once you find a common denominator, put it under a common denominator.*0454

*The denominators go away.*0460

*They are equal, therefore, the numerators are equal.*0461

*We concern ourselves only with the numerators.*0464

*Therefore, we have 12 is equal to, I can multiply this all this out here.*0476

*We have ax² + 9a + bx² + cx - 2bx – 2c.*0482

*12 =, I’m going to take care of the ax² bx².*0506

*It is going to be x² × a + b, that takes care of the ax² + bx².*0510

*We will do the x terms.*0518

*The x terms, I have c - 2b that takes care of the x terms.*0520

*I have + 9a - 2c, 9a - 2c that takes care of the number terms.*0528

*Now I equate coefficients.*0540

*Over on the left, there is no x² term.*0542

*Therefore, a + b is equal to 0, 0x².*0545

*Therefore, I have the equation, a + b is equal to 0.*0549

*On the left, there is no x term, therefore, it is 0x.*0553

*Therefore, c - 2b is equal to 0.*0557

*The number 12 is that one.*0561

*Therefore, I have 9a - 2c is equal to 12.*0565

*These three, these three equations and three unknowns is what we are going to solve for a, b, and c.*0571

*Let us go ahead and do that next.*0576

*I have got a + b = 0, I’m going to write them this way.*0582

*C - 2b = 0 and 9a - 2c = 12.*0586

*I presume that most of you are comfortable with solving two and three, sometimes four equations, and that many unknowns.*0596

*But I will go through the process anyway, it is not a problem.*0603

*It only takes a couple of minutes.*0605

*Here a = -b, here c = 2b.*0607

*I’m going to put this a and this c into here and solve for b, and then put the b’s back and find a and c.*0614

*I have 9 × a which is - b - 2 × 2b which is c equal to 12.*0621

*I have -9b - 4b = 12.*0633

*Sorry, this looks like a 13, this is a b.*0641

*I have got -13b is equal to 12.*0644

*Therefore, I find that b is equal to -12/13.*0648

*That takes care of b.*0654

*Now I go ahead and put that over here.*0655

*I find that a = - a -12/13 which means that a is equal to 12/13.*0657

*Of course, c is equal of 2 × b which is -12/13.*0669

*Therefore, c is equal to -24/13.*0675

*Now I found a, b, and c, I put them back into my original decomposition.*0680

*Remember, we had that our original 12/ x - 2 × x² + 9 is equal to, our decomposition was a/ x - 2 + bx + c/ x² + 9.*0686

*Therefore, we just put it in.*0706

*I have a is 12/13, this is going to be 12/13 / x - 2 + b which is -12/13 x + c which is -24/13 / x² + 9.*0707

*There you go, this is our partial fraction decomposition.*0735

*This is the first part, we did our partial fraction decomposition.*0742

*Now we actually want to integrate this.*0745

*The integral of this is going to be the integral of this.*0747

*The integral of this is going to equal the integral of this + the integral of that.*0751

*That is it, just work your way through.*0755

*Let us see what we have got here.*0759

*Our integral, which I will just call int, is equal to the integral of 12/13 / x - 2 dx*0766

*+ the integral of -12/13 x - 24/13 / x² + 9 dx, which is going to end up equaling 12/13*0779

*× the integral of dx/ x - 2 - 12/13 × the integral of x/ x² + 9.*0809

*Just separating this thing out -24/13 the integral of, this is x dx, sorry about that.*0824

*I always forget the dx, for all these years, I still forget it.*0836

*x² + 9.*0841

*We end up getting the following.*0846

*We end up getting 12/13 × the natlog of the absolute value of x - 2 - 12/13 × ½ the natlog of x² + 9.*0847

*This ½ factor came from the fact that this is a u substitution.*0866

*I let u equal x², therefore, du = 2x dx x dx.*0870

*Bring the 2, u substitution, I will let you work that out.*0876

*This one is going to be -24/13 √9.*0882

*I will tell you where this came from in just a minute.*0890

*1x/ √9 + c, there we go.*0894

*The last integral, this one right here, where did I get that?*0907

*Here is where I got that.*0914

*For the last integral, we use the following formula.*0916

*The integral of 1/ x² + a² dx is equal to 1/a × tan⁻¹(x)/ a.*0927

*That is the form that we use for this because often, when we do partial fraction decompositions,*0943

*especially when we have quadratic factors in the denominator, we often end up with integrals that look like that.*0947

*Some dx which is just the 1 dx over here/ something x² + something else x² + something else².*0953

*It is the general integral that keep showing up.*0966

*We went ahead and we are just going to use the formula for it.*0970

*1/a × tan⁻¹(x)/ a.*0972

*Let us do another example here, this time we want to evaluate the integral of x – 4/ 4x² + 4x +5.*0981

*We take a look at this and we realize that the denominator cannot be factored any further.*0990

*This is definitely an irreducible quadratic factor.*1007

*4x² - 4x + 5 is irreducible and it is also the only factor.*1014

*Because it is the only factor, there is no partial fraction decomposition.*1039

*The rational function itself, it is the partial fraction decomposition.*1043

*It is a partial fraction decomposition that is composed of just one term, x - 4/ 4x² – 4x + 5.*1048

*There are no other factors in the denominator, for me to actually expand and do what I have done in the previous problems.*1055

*Here we are going to show you a general procedure for how to handle a quadratic in the denominator that is the only term.*1061

*Here is how we do it.*1070

*We actually are going to complete the square in the denominator.*1074

*We handle the situation, again, this is a general procedure,*1079

*anytime you have a quadratic in the denominator that it is the only factor.*1082

*It is a single only factor.*1086

*We handle this situation by completing the square.*1089

*Something that you have done thousands of times in algebra, by completing the square in the denominator.*1096

*I got to tell you the technique of completing the square is something that comes in handy so often,*1107

*and so many other branches of mathematics.*1114

*We are working just with the denominator.*1120

*The denominator, we got 4x² - 4x + 5.*1125

*I’m going to go ahead and factor out the 4.*1134

*This is going to be 4 × x² – x.*1136

*I will leave a little space for something that I add, + 5.*1142

*I take half of the second term which is -1/2 and I square it.*1146

*This is going to be + ¼.*1149

*Since I added 4 × 1/4, I added 1, I'm going to subtract 1 from this expression to retain the equality.*1156

*I'm going to write this as 4 × x – ½² + 4.*1164

*This is just 2² × x – ½² + 4.*1174

*Again, this is just mathematical manipulation, nothing strange happening here.*1181

*This is 2² × something², I’m going to put them together and take the squared out.*1185

*This is going to be 2 × x - ½² + 4.*1191

*I'm going to multiply, I’m going to distribute the 2 in there.*1202

*This is going to end up being 2x - 1² + 4.*1204

*Now I have that, this is my denominator.*1217

*I just changed the way it looks.*1226

*We have the integral of x - 4/ 2x - 1² + 4 dx.*1235

*Now I’m going to subject this to a u substitution.*1247

*Let us do this in red.*1254

*I’m going to let u equal 2x – 1.*1255

*I’m going to let du = 2 dx, that means dx is equal to du/2.*1263

*I have taken care of that.*1274

*Over here I’m going to actually solve for x.*1278

*It is going to be x is going to equal, because I want to also deal with this x on top, in terms of u.*1280

*x, when I solve this equation, we said u = 2x - 1 so x = u + 1/ 2.*1291

*When I put all of these in here, we have the integral of u + 1.*1299

*I’m going to write it as ½ of u + 1, that is my x, - 4/ u² + 4 dx is du/2 which = ½.*1316

*I’m going to pull this ½ out.*1338

*The integral of ½ u + ½ - 4 which I’m going to write as 8/2 / u² + 4 du*1339

*= ½ of the integral of ½ u - 7/2 / u² + 4 du.*1363

*I’m going to pull the ½ here, put it here, it equals ½ × ½ the integral of u - 7/ u² + 4.*1374

*So far so good, that = ¼ × the integral of u/ u² + 4 du - 7/4 × the integral of 1/ u² + 4 du.*1389

*Now these integrals, I can handle.*1413

*Remember, again, u = 2x – 1.*1416

*Let us actually write this again.*1425

*We said that it equal ¼ × the integral of u/ u² + 4 du – 7/4 × the integral of 1/ u² + 4 du.*1437

*That gives us ¼ × ½ × natlog of u² + u – 7/4 × ½ tan⁻¹ of u/ 2 + c.*1455

*This ½ term, that one comes from the fact that I do a second u substitution on this.*1485

*If I call that one v, v = u².*1494

*This ½ comes from the fact that this is that formula, it is 1/a u² + 2².*1502

*Remember that formula that we just did.*1511

*We said that the integral of 1/ u² + a² = 1/a × tan⁻¹ of 1/a.*1515

*Here this is a² which means that a is 2.*1525

*That is where this one actually comes from.*1530

*Sorry this looks like a u, a² + 4, this is the u and this is the 4.*1540

*There you go, again, this is a general procedure that you can use whenever you have an integral where you have only one factor.*1550

*Or the rational functional only has one factor and it is an irreducible quadratic.*1560

*You complete the square on that thing and then you use a u substitution to do what we just did.*1564

*It will always work.*1571

*Let me actually write that down.*1577

*This problem offers a general procedure for dealing with integrals of the type,*1581

*on the very attractive integral sign, it is b/ ax² + bx + c dx.*1611

*Anytime you are faced with an integral that looks like that, you can run this procedure.*1627

*Complete the square and then solve the integral.*1631

*Now let us deal with our 4th and final case.*1637

*We just did irreducible quadratic factors that are non-repeating.*1640

*What if we have repeating quadratic factors?*1644

*That is actually going to be the same thing.*1646

*It is just more terms in your partial fraction decomposition.*1647

*Case 4, it is where our numerator/ our denominator is our rational function.*1654

*It is where our denominator has repeated irreducible quadratic factors.*1664

*In other words, it is a quadratic factor of the form ax² + bx + c raised to some power.*1686

*It itself might be squared, a quadratic factor might show up two times, three times, four times.*1713

*It is called the algebraic multiplicity, the multiplicity of the factor.*1719

*The partial fraction decomposition of something like this is as follows.*1740

*You have a1 x + b1/ ax² + bx + c + a2 x + b2/ ax² + bx + c²,*1743

*and so on, until you get to a sub n x + b sub n/ ax² + bx + c ⁺n.*1770

*In other words, whatever n is, you are going to have that many.*1783

*You are going to have that first power, second power, all the way up to nth power.*1789

*You are going to have all of these linear factors up on top.*1794

*You have all of these coefficients to find.*1799

*Let us do an example, I think it will make sense.*1805

*Same exact thing that what we did for the repeated linear factors, just have them keep showing up to the nth power.*1809

*Evaluate 2/ x × x² + 6².*1820

*In this case, this quadratic factor is irreducible, x² + 6 cannot be factored.*1824

*It itself is raised to the second power.*1829

*The denominator is already factored, we do not have to do that.*1833

*It is already factored, therefore, 2 divided by x, x² + 6².*1846

*We have a linear factor that is the x, it becomes a/x.*1855

*We have a quadratic factor raised to the power of 2.*1860

*We are going to do bx + c/ this quadratic factor to the first power + dx + e, the quadratic factor raised to the second power.*1862

*Our partial fraction decomposition involves one term where the quadratic factor is to the first power*1879

*and the second term where the quadratic factor is to the second power.*1886

*As many terms, all the way up to that many powers.*1890

*If this were a 3, we would have fx + g/ x² + 1³, and so on.*1892

*This is it, now we are going to find the least common denominator on this side.*1899

*Let me actually write this out.*1909

*The least common denominator here, be very careful, the least common denominator is not this × this × this.*1912

*It is this × this because this factor is already contained in that.*1917

*Our lowest common denominator is x × x² + 6², and whatever is on top.*1926

*Be very careful, you are used to having the least common denominator, just multiply the denominators.*1940

*Here, because the factors are repeated, this does not need to be this × this × this.*1945

*It does not need to be that way, this is already contained in that.*1951

*It is just this and this.*1953

*Therefore, here what we need to do is we need to do a × x² + 6² + bx + c × x² + 6, only once.*1955

*And + dx + b × x/ x × x² + 6².*1981

*This denominator is the same as that denominator, which means the numerator is equivalent to the numerator.*2003

*Now we expand this numerator and that is what we are going to do next.*2009

*2 is going to equal, when I multiply all this out.*2016

*That is fine, it is just algebra.*2025

*ax⁴ + 12ax² + 36a +, it is going to end up being bx⁴ + cx³ + 6bx² + 6cx + dx² + e ⁺x.*2030

*When we combine terms, we have an x⁴ term.*2070

*This is going to be a, it takes care of that one, and a b.*2077

*It takes care of the x⁴ terms.*2084

*There is an x³ term.*2087

*The only x³ term is that one, c.*2089

*There is an x² term, 12a + 6b + d.*2099

*Make sure you get all the terms.*2113

*There is an x term, 6c + e.*2116

*There is a number term, 36a.*2131

*All of that is equal to 2.*2138

*We just set things equal to each other.*2143

*The equations that we get, in other words, a + b is going to be 0, c is going to be 0.*2146

*12a + 6b + d is going to be 0.*2153

*6c + e is going to be 0.*2157

*36a is going to equal 2, that is what we get.*2160

*We are going to get a + b = 0.*2167

*We are going to get c = 0, we are going to get 12a + 6b + d is equal to 0.*2172

*We are going to get 6c + e is equal to 0.*2184

*We are going to get 36a is equal to 2.*2189

*This gives us that a is equal to 1/18.*2193

*That takes care of that.*2198

*Let us go to the first one over here.*2201

*We use the equation a + b is equal to 0, which means that a = -b which means that b = -a which means of b = -1/18.*2203

*We already know that c is equal to 0.*2222

*We have taken care of that.*2227

*Now we have 6c + e is equal to 0.*2229

*c is equal to 0 so I get 0 + e is equal to 0, which means that e is also equal to 0.*2237

*Now I’m going to use this equation.*2247

*12 × 1/18 + 6b, 6 × -1/18 + d is equal to 0.*2253

*When I solve this, I get d is equal to -6/18.*2267

*I’m going to leave it as -6/18, instead of reducing that.*2273

*Therefore, our final partial fraction decomposition, our original function was 2/ x × x² + 6².*2276

*We said that was equal to a/x + bx + c/ x² + 6 + dx + e/ x² + 6².*2289

*Now we have a, b, c, d, and e.*2304

*Our decomposition is 1/18 / x + -1/18 x + c which is 0/ x² + 6 + -6/18 x + e which is 0/ x² + 6².*2308

*This is our final partial fraction decomposition.*2336

*That is just the decomposition, that is not the answer.*2340

*We still have to integrate this thing.*2342

*The integral of this is the integral of this because these are the same.*2346

*The integral of this is the integral of this + the integral of that + the integral of that.*2351

*Nice and simple, the rest is just using all the techniques that we gathered so far.*2357

*My final answer, our integral is going to equal 1/18 × the integral of 1/x dx - 1/18*2362

*× the integral of x/ x² + 6 dx – 6/18 × the integral of x/ x² + 6² dx.*2378

*This is equal to 1/18 × natlog of the absolute value of x - 1/18 × ½, because of that u.*2403

*u = x² + 6, du = 2x dx, x dx = du/2 × natlog of x² + 6 – 6/18 × the integral of x/ x² + 6² dx.*2417

*How do we handle this?*2444

*We handle it this way, we do we a u substitution on this.*2448

*We let u = x² + 6, du = 2x dx, du/2 = x.*2452

*It is essentially the same thing that we did here, because of that extra x² part,*2467

*I thought it actually do the u substitution, x dx.*2470

*We have ½ × the integral of u⁻² du which is equal to ½ × u⁻¹/ -1 which = -1/ 2u which is equal to -1/ 2 × x² + 6.*2478

*Our final answer is = 1/18 × natlog of x – 1/18 × ½ × natlog of x² + 6*2509

*– 6/18 × -1/ 2 × x² + 6 + c.*2531

*There we go, partial fraction decompositions are very tedious.*2547

*They are algebraically intense.*2552

*There are plenty of places where you can make a mistake but conceptually I do not think it is all together that difficult.*2555

*You just have to keep track of everything.*2560

*But this is calculus, you are more than accustomed to that by now because the problems just are,*2562

*by nature, sort of long and detailed.*2569

*Thank you so much for joining us here at www.educator.com.*2574

*We will see you next time, bye.*2576

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