 So I guess last time we did some partial fractions, but maybe not enough. Is that right? Alright, so let me just do a couple of examples. So the situation is where we have something, we have something like this. Oh no, this one's easy, isn't it? Oh well, say we have something like this anyway. Yeah, that was a stupid one. Oh well, if we have something like this, then what do we do? We're also here. Why don't we do fractions? So let's do that first. 26, what? I don't know if you want to change it. It says 41 or 26. Okay, I don't know what that means, but alright. Okay, 26. The other 26 or 41, it tells me more. I don't know if I need it. 26 works, but it also tells me 41 does. 41 works too. So, can you change it? Probably central. So when you stop this, do you need more time? Ask you to actually do me. Do they still need more time? Okay, so most of the people like this option, something like 75%. And then this is number two. Not too many people like this one. In fact, all of them work. It's just that if you do the substitution, so let's do the substitution first. So here if I make the substitution, obviously I want to let u equal x squared minus 1. I don't know if that's obvious to you, but it's completely obvious to me. And then du through x dx, and I almost have dx sitting around there. I have that 3, but so what? So if I do that, then I have x minus 3 x squared minus 1 dx is, so then this will be, I put this part and then I have the 3 left over. So actually let me leave that in. And now this I can do easily with this substitution. So this is 1 half du because x dx is 1 half du. 1 half du on the top and du on the bottom. And then here, this is one I should know. Well, I guess this doesn't matter, right? Because now here I have to do this 3 over x squared minus 1. This is not 1. Now I can make a trig substitution, I suppose. But I don't want to do this, right? To do this 1 over x squared minus 1, I have to make the substitution of secant squared. I don't know. This is ugly. Is this ugly? Yeah, this is ugly. Right? I mean, this I would have to do by trig substitution. But I don't really want to go here. Is this ugly or am I just asleep? It's ugly, right? So I make a secant and I make a secant squared integral of, well, that's cosine squared and then I have to do 1 half cosine. Yeah. You still can't plug in the u to that? Here? Yeah, where minus 1. No. No. So if I do 3 over u, then I have left over. I need an x. So I need some x here. To change du over dx, this would be 3 over du over 2x. That's not so good. Yeah. So I can't just plug in the u because I have to deal with the dx. So this I would just leave and now I'm kind of stuck. I mean, here if you have this, you can do this. So I can use a trig substitution that's even squared. X is mu is, let's call it, no, sorry, even squared is x and then I get the integral of a cosine squared and then I turn that integral of a cosine squared with a half angle formula and it works. But actually, maybe I shouldn't be sad. I should be like a sweater. It's a sweater. So it's here. Anyway, you can do it and it will eventually get there, but it's not the most efficient answer. So formally the correct answer is either a or b and c together and depending on how much of a masochist you are, it either doesn't matter or it does. I mean, it matters in the sense that you've got to kill yourself to do something that's relatively easy. So let's do it the easy way now. So the way that most of you want to do is we just use partial fractions. So just to remind you, I just need to solve a little algebra problem a over x plus 1 plus b. This is what I wrote. Over x plus 1 equals x minus 3 over x plus 1 times x minus 1. So I want to solve this little algebra problem and figure out what a and b are. And I can either do this by cross multiplying and getting two equations that lead me to that. So this is the same thing as if a times x plus 1 plus b times x minus 1 equals x minus 3. Right? Because if I find a common denominator and I get that. And now here, I can either say, alright, looking at this, a plus b has to equal 1 because the coefficients match. So I'm going to do this two ways. Did you do it both ways? No? Okay. I don't know. So I'm going to do it both ways. You choose which way you like. They're the same to me. So either I can say, alright, if I clean this out, I have, there's a 1 there, a x plus a, but I'm going to not worry about that. I'm going to leave the x off in fact. a plus b has to equal 1 because the coefficients of x, a plus b, have to match. And then the coefficients of the constants, a minus b, have to equal minus 3. So this is from the x and this is from the units. And then I solve, that's a 3. It's just a funny looking 3. And then I solve those simultaneously. So if I add those things together, I get that 2a equals minus 2. And so a minus 1. And since a plus b is 1, that means b is 2. So that's one way I can do it. An equivalent way I can do it is I can look at this equation and I can say, okay, this equation has to be true for all x's. So in particular, this equation has to be true. This equation has to be true when x is 1. If x is 1, then this is 0 and it's gone and it's out of my face. So that means that a times 1 plus 1, so that's 2a equals 1 minus 3. And so that means, again, a is 1 because 2a has to be negative 2. And if x equals negative 1, which kills this piece, then I have here negative 2b equals negative 1 minus 3, negative 4, so that means b equals 2. These are the same, just done in a slightly different way. I mean, what's making them work is exactly the same fact, which is that this equation has to hold for all values of x. So either you can equate coefficients or you can pick suitable values of x and whichever one makes you happy is fine because they're really the same. Alright, so in both cases, in both cases we wind up with something fairly easy. So that tells me that a is negative 1 and b is 2. So that means that the integral x minus 3 dx over x plus 1 times x minus 1 is the same as the integral of negative dx over x minus 1 plus the integral of 2 dx over x plus 1. And this is really easy, right? So this is, look at it, this is the, if I make the substitution, u equals x minus 1, then I get 1 over u. So this is plus 2, the log of x minus 1. And here I make the substitution, u equals x plus 1, so I get 1 over u. So this is plus 2, the log of x plus 1. So these are really easy once you get the hang of it. The hard part, so the hard part with partial fractions actually, as in most calculus, is not the calculus but the algebra. But this is pretty straightforward. Any questions on this at all? Then why did you make me do it? Okay, so let's just go on. Let me do another one. So unfortunately, and I think this is what she did at the end of the class that she told me confused you all. Maybe all the factors are not linear. Maybe I have a power on the top or the bottom or something like that. And I can't factor it into linear factors because I can't almost factor everything into linear factors. So the problem here is, well let me just do an example still. So say I have the integral of 3x squared minus x. Let's get more of these out in advance. So 5 over x minus 2 squared plus 1. And there's a dx here somewhere. And again, it should be pretty obvious that what I really want to do here is partial fractions because, so now I want to be able to split this up. The problem is, so let me actually move over to the next board. So I have this problem 3x squared minus x plus 5 x minus 2 squared x plus 1. And I want this to be equal to something over x minus 2 squared plus, I'm going to skip B because I'm going to use it in a second, over x plus 1. But this won't actually work because I'm going to be missing an x. This only gives me a linear term in x and it doesn't give me full choice on the x term. It will be full choice on the constant term. The constant in the x term are related by that. So I need a slightly more complicated numerator here so that I have the full freedom to vary the coefficient of x. Well this is really the coefficient of x squared and the coefficient of x. Because when I cross multiply, this one gets a quadratic term, this one gets a linear term and I lose out on the freedom. I need three things to vary because I have a quadratic polynomial or a cubic polynomial so I need three degrees of freedom. So the general rule is, or at least some people say it slightly differently, when you take your factors you always put the numerator. So the numerator is always a polynomial, one degree lower than the denominator. And then maybe I can split this up. Now those of you that may have seen this before or the way they describe it in the book, they say, okay, take A over x minus 2 squared plus B over x minus 2 plus C over x plus 1. Those are the same which is slightly different choices of A, B and C. The advantage of doing this is this unifies the two second cases. Any questions on that at all? Yeah. Does it matter which letter is assigned the x value? No. Would you like to take this be the B and this be theta? I mean it doesn't matter, now you can't read it so that's probably bad. The letter, I mean we can put a smiley face here and a star here. It doesn't matter because it's just a placeholder. So use whatever letter you want. But then once you start you consistent. You understand? So what matters is that the thing on top of the quadratic term is a linear polynomial and the thing on top of the linear term is a constant. It's like a facing star. And okay, if you like to find a facing star you keep it. Alright, so now we can use the same trick. We just cross multiply and come up with something. So when we cross multiply we're going to get smiley face x plus star times x plus one plus c times x minus two square and that has to equal that three x square minus x plus five. And so now we need to figure out what smiley face star and c are. And again we can use either method. It's just a quite coefficient system. So if I quite coefficient, well, yeah let's say quite coefficient. So otherwise I'll have to choose two values of x. It doesn't matter. So here if I multiply this junk out I get smiley face x square plus smiley face plus star times x plus star plus. So I'm just distributing this. It's confusing because I'm not using actual letters. And then here I have to square this out. So this squares out to be x squared minus four x plus four c and when I multiply that by c. So this is cx squared minus four cx plus four c and that has to equal three x squared minus x plus five. And so from there now we can match things. So the x squared gives me a three. So I have smiley plus c equals three. And the x's tell me that smiley plus star minus four c equals minus one. And then from the unit's terms I get the star plus four c plus five. I have to solve these things all together. So if I add these together I get, so adding this one and this one give me smiley face plus two stars equals four. Yeah, good question. Okay, so let's go through this process again. I look at this equation here. I cross multiply and I get some long mess. Now what has to be true, this is what's in front of an x squared. In front of an x squared I have one smiley face, I have a c, I have a three. And this is what's in front of an x. In front of an x I have a smiley added to a star, I have a minus four c, I have a minus one. And this is what's in front of a one. In front of a one here, well that has an x. In front of a one here I have a star and I have a four c. On this side I have a five. So let's get through my three equations. And now I just want to say, if I have three numbers that when I add up in this way, if I have three numbers they have to satisfy these three equations. And now I want to solve these three equations. What should I do with the other way? If I do it the other way then it tells me c right away. So let's look at the other way. So you can solve these three equations and probably get the same answer I wrote the other way. The other thing I can do is I can say this has to be true for all x's. If x is one and this messier is zero, x is minus one, then I have smiley plus star times zero plus c times minus one minus two square equals something like three plus one plus five. So this is one and so this is nine. So I get nine c equals nine. Well that's good. So that tells me right away that c is one. Now that I know what c is, then I can go back and figure out what smiley and star have to be. If I take x to be two, well that gives me a relationship between smiley and star. I'm actually going to mix these two methods. So I know that c is one which makes my life a whole lot easier because I know that smiley plus one is three from this line here. Well since smiley plus one is three that means smiley is two. And since smiley is two and here I know that c is one so star plus four c is five. Star plus four c is five so that means that star is one. So now mixing the two methods actually turned out to be the most efficient way. You can of course solve these by adding and subtracting and bushing the graph. But it works. And you can of course use this method exclusively as well. Yeah. Would you like to have d over x minus two? d will be zero. So let me finish this and then I'll address that. So are we okay on this? So use whatever method works as long as it's legal. It's fine. I'm mixing the two methods. They're the same method so it doesn't matter. This all comes down to the fact that if two polynomials are equal they have to be equal at every x value and their coefficients have to be equal. But if two things are equal they have to be equal so that means they have to be the same. So I can use their various aspects to identify them. If you're trying to identify your mother you can listen for her voice or you can look at her face. They're the same. They're both ways of identifying who your mother is. Okay. So anyway, so we have this. And so now this problem becomes easy somewhere. I guess it will be easy over here. So that means that the integral of the x squared minus x plus five dx over whatever the drunk is x minus two squared x plus one is the same thing as the integral plus two is what? Smiley is two, two x plus one over x minus two squared dx c is one dx over x plus one. And then this guy we can do a substitution and this guy is a log so away you go when you stop it. So let me just not do that. Now let me come back to somebody over here. I think in the blue, right? Yeah. So her comment about why don't I use d instead. Because it's the same. Let me use d on that point. So if you like, if it makes you happy, I could write this in a different way. I could say instead that x squared minus x plus five x minus two over x minus two squared, well let me get into those. B over x minus two c over x plus one. I could do that too. Is that what you meant? No? Forget it? So I can do this as well. It's the same. Because if I put these things together then my b will just be some combination of Smiley's and the stars. Right? Because so a is the same as Smiley and b is the Smiley and the star which is b plus a, b plus minus two a is the star. Oh yeah. But it's still the same thing more. So the a is the star. I mean b is the Smiley and a minus two b is star. So that doesn't matter. And if you want to put more stuff up here, if you want to put d's and e's and q's times x's and x's q's, it's fine but they'll all be zero. So wipe off. So yeah. What else would you do? Okay. So if I tried to do x minus two and x plus two and I wouldn't have a way, you mean you would like this one to be like that? Is that what you're asking? Okay. So this is a good question and it's something that confuses people all the time. Imagine that I did that. So imagine that I wrote. Well okay. So let's do a simpler example. Suppose I did that. Suppose I wrote a over x minus two plus b over x minus two. How can I get an x squared out of it? I can't. But that has to equal something with an x squared. Right? Because I have an x squared on the top. I can't get an x squared out of it. Because when I cross multiply, I'm going to have a times x minus two plus b times x minus two. Well you're not going to get a square. So it's not going to happen. So I need to have something squared. So I can square this and then when I cross multiply, the b will get the x squared term. But somehow I'm going to have to pick up the x squared term. So the general situation is that whenever I have something for the factors, well this is not as bad. Whenever I break this down, the thing on top is always a polynomial of just one degree lower. That's the thing on the bottom. Now sometimes you might have to, like in this example actually, you might have to split it up. Because when I make my substitution, it's not going to quite work. But I can split that up. You see that? Right. So if I have the integral of x minus two cubed times x over, I don't know, x. So I want to split this up. But I'm going to write this as a over x minus two cubed. I can either write this as three things where I do a two, a three, or one. Or I can just write it as one thing, like that. And since I put an integral there, I better than break here. Yeah. Yes? Are you solving by elimination? So, okay. I am solving by elimination, yes. So I can solve this by elimination. But I have extra information that I might as well use. I don't like solving three equations and three unknowns. Because I combine this and I combine that. And now I get two equations and two unknowns. And then I solve the two equations and two unknowns. And then I go back and find the other one. It's easy for me to figure out what c is not using this. So by looking at the equation here, that if x is negative one, this term is gone. When x is negative one, this thing is zero. And then I have three c, negative nine c equals nine. So by the judicious choice of x, I can make the equation simple. I'm saying sometimes one method is faster than another. So what I don't like to do is say, always use this method because it's always better. It's not true. Sometimes one method is better. Sometimes another method is better. Sometimes the mix of the two methods are better. There are two methods. They're the same really. But they seem different. You have a different feel. And you use whichever one works best at the time. You can, I can do this. I can forget about the other method entirely. And just solve this by itself. So if you're confused by the other method, forget about that. Just solve these three equations and three unknowns. We get the same answer. I guarantee it. But it's slow. It sucks. But you're welcome to do it. This other method isn't going to work so well here because I can kill the x squared minus two by substituting x equals two. But then I'm left with something involving both smiling and star. So I need to make another choice that will not kill all three terms and put that together. So instead of doing that, it's going to be these two methods. They're all the same things. So you can do it. Okay. Other questions about this? We will, I mean, it may seem like, I mean, this notion of partial correction comes back. So outside of doing integrals, it still comes up in other areas. So let me stop, stop with this guy. Let me point out one other thing that will, more algebra. So really today is an algebra class. Suppose that I have something like, do the same integral, x equals minus four x squared plus one over x squared minus one. So let me figure out the integration part. So suppose I want to do this. What should I do? I write this as 26 and it's 41. Both are supposed to work. You have to wait a minute. I have a lot of people. So I mean, I'll stop there unless you need more. Wait. Okay. If you'd like a coin just let me have one of them. Okay. So you still need more time? No? Yes? Yeah. So by overwhelming majority, you think this won't work. And you write, why won't this work? I can't get an excuse. So what do I do? I do what? I divide it. So it does come from the denominator. I mean, I can't start with partial correction but do something first. So the problem here is that the power here is three and the power here is two. If you prefer I could write this as x squared minus one. The power here is three and the power here is two. The partial fractions is not going to work. I have to clean it up a little bit first. So this is the same thing minus four x squared plus one x squared minus one. And so what I have to do is reduce the power of the numerator. So I want to get that down to being an x squared. So I mean actually, yeah. So I need to do division. I need to say if I multiply the bottom by x, I'll get an x cubed that I can subtract off. So I have to do long division here. I don't have to do anything exactly this form but okay. So leave room for x. And probably a lot of you forgot how to do long division with actual numbers. How many people remember how to do that? If I want to do like 15 into 209 and 61, people remember how to do this from like third grade. It works exactly the same. So when I'm dividing 15 into 2096 I say okay, well one goes into two. Two times, well that's too big. So I better use one time. So I put a 15 here. I put a 1 in here. I subtract. I get 59. And I say how many times is 15 going to 59? Five times, no. Four times, no. Almost. Three times. And then I put a 45. Three. And I subtract. And I get 146. And then I say 15 into here. Maybe nine. No. Made. And so on. I need to do this. Nine. Nine or 45. You're right. Okay. So nine works. And when I do nine into here I get 90 plus 45 is 135. One. I'm going to bring down the one. I'm already tired. Now it's so close I can finish, right? So if I go 15 into here it goes four times, five times. Seven. Seven is too big. Six. If I do six then I get 60 plus 30 is 90. Seven. One out of five and I have a remainder of six. I'm actually going to write it as a remainder of six rather than writing six 15s. But if you prefer you can write it as six 15s. Okay. Here the trick is exactly the same. I look at the x square. I say how many times is it going to x cubed? It goes in exactly x times. And so if I close. What am I doing? So it goes in x squared goes into x cubed exactly x times. And then x times x squared minus one is x cubed minus x. And I subtract. Which is exactly what I did here. 15 goes into 20. How many times? One time? One times 15 is 15. Then I subtract this part. And then I bring down. So let's go. So here I subtract. The x cubed is gone now. When I subtract here I get a four minus four x squared minus x plus one. Plus I subtract the minus. Yes thank you. And I can do this one more time. I can say how many times will this x squared minus one go into minus four x squared plus 12? Well x squared goes into four x squared minus four times. So then I multiply back. Just like I did over there. Minus four times x squared is minus four x squared. If this term doesn't cancel you screw it up. And then minus four times minus one is plus four. Four. And then again I subtract. Those guys cancel. X minus zero is x. And this guy I mean one minus four is minus three. So then, well I can't put that one up. Okay. So then this problem becomes, so that tells me, oh I guess I need the remainder. This is my remainder. If I try and divide x squared plus one into this I will need to divide by x rather than multiply by x. So that means that this process has shown me that, where is my problem? That x cubed minus four x squared plus one over x minus one is the same thing as x minus four plus x minus three over x plus one. x minus five is four. And so if I'm integrating that, so if I want to integrate this, I integrate this and I integrate this. And this one I did already to start at the plus. So I chose this on purpose to make sure it would come out. So now I can do this by partial fractions. Do not do this by partial fractions because I only did it. It's right there. For partial fractions to work, the degree of the numerator could be one less because otherwise you can't pick up the answer term. So sometimes they have to be arranged. Okay. So we finished this with peaks of doing integration. So we now know as much as I do about how to integrate. But now there's some other variations on that which we'll talk about in the next portion of the class.