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Makayla B.

July 17, 2021

In the first example you say that the 1 becomes -1 but I thought it would be a positive 1 because you would fill in the blank space when you are subtracting with a zero and positive 1 minus 0 is positive one?

Oregon State University

University of Michigan - Ann Arbor

University of Nottingham

02:07

Fangjun Z.

If $f(x)=x+\sqrt{2-x}$ and $g(u)=u+\sqrt{2-u},$ is it true that $f=g ?$

0:00

Felicia S.

Dungarsinh P.

Jsdfio K.

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well, the rational functions that we've been dealing with and that is by long division. So again, we know Long Division is another way Thio integrate. So that's our goal is to integrate a rational function, which is in the form of ffx divided by G of X. But how do you know when to use long division verse? Partial fractions? We've talked about using partial fractions when the f of X degree. So when the numerator degree is less than that of the G of X degree, we use long Division. When the opposite happens. When the degree on FX is greater, then the degree gov. Here's the situation where you use long division. What does Long Division look like? Let's say we have f of X divided by G of X. We use long division, which means we divide we say g of X divided by f of X. We will go over lots of examples of how to do the long division before we get into actually solving the integral. So we do G of X divided by ffx. What this is going to give you is you're gonna have some function on top. We'll call this h of X, so you'll have some function on top, just like with regular division, and then you'll probably have a remainder and you'll have a remainder. So let's say we have f of X divided by G of X, and we do this long division. The division is equal to your function, which is h of X. So whatever is on top, plus your remainder which will call RM. Remainder could be a number could be a function divided by G of X. Okay, so your division is going to be equal to whatever you end up with on top, plus whatever your remainder is divided by G of X. If we're going to integrate this, we can add integral side. The integral of ffx G of X would be the integral of this, which would be equal to the integral of H of X because in a world's could move through places, Reminder integral of remainder over G of X. This now might require partial fractions, so it could be a two step part where you use both partial fractions and long division. So the key part to remember we're going to talk about it one more time long division is used when your ffx degree greater G of x degree. This is when we use long division. Partial fractions is used when your ffx degree is less than your G of X degree. Once you've established that we've talked about partial fractions. Let's say we have ffx divided by G of X. When you do long division, this is equal to G of X, and then you put ffx inside. You do the division, you're going to get some function up here. You'll also end up with a remainder. This is equal to how you write this. In a sense of an equation is that ffx over g of X, or whatever fraction you had is going to be equal to your h of X plus whatever your remainder is divided by G of X. So this is gonna be the form that all the answers for long division should be in, and then you can add the integral sign in when needed. So we're just going to see a couple of examples of how to set up the long division how to get the answers, and then we're going to do the majority of the examples with actually solving the integration all the way through how to do long division. So let's say we're doing X squared minus X divided by to execute minus two X squared plus one. So this is just the general process, and then we're going to do four more examples that actually use this technique to solve integral. So to do long division you want to see how do you get two X cubed from X squared? So what goes on top is what you need to multiply x squared by to get to excuse So we needed to and we need an X and the new multiply. So two x times two X squared is to execute you multiply two x by x so you have minus two x squared. You don't have anything over here. Then you subtract this so two X cubed minus two X cubed is zero and then you get minus two x squared plus two x squared, which is also zero, and we just bring down our one. But you have to bring down your one is a negative one. Don't forget that we are subtracting this whole equation. Okay, then. So now there's no other X square cannot go into negative one anymore. So now what we have is that we're done. So now this is equal to how you would write. Your answer is two X So whatever is on top minus this one is considered your remainder one divided by X squared minus X your original function you're dividing by. So this would be your final answer when you're doing Long division. So now we're going to use long division toe, actually, help us solve these equation solved these in a gross and just a reminder. We use long division when the degree of the numerator is greater than the degree of the denominator. So the degree on top is greater then whatever degree you have on the bottom of your fraction. So this isn't the most mathematically correct way to write it. But if saying f of X and G of X is getting you confused, the top degree used to be greater than the bottom degree to use long division. And then we know when we use partial fractions. Look at this one. So the degree on the top is four. The degree on the bottom of the fraction is three. That means we should not be using partial fractions. So we're going to go ahead and use Long Division. So for long division, this is going to be equal to. So let's we need to rewrite our integral. So we're going to pull out the equation, and we need to replace this equation in the integral with something else. So this is equal to the This is equal to y cubed minus y squared plus y minus one and then too wide to the fourth using long division. So we need to multiply by two y because too wide times why cute gives me to y to the fourth. And then this is like plus zero plus zero plus zero. But we still have to go through all of these terms. So to I minus y square, we get negative to Why cute plus two y squared minus two. Why? And we have to do subtraction. Don't forget, we're subtracting everything here, so we do need to distribute that negative. So too wide of the fourth minus two y to the 4th 00 minus a negative. Two y cube is plus two y cube. We have a negative two y squared, plus a to why? Now you have to look and see if you can continue. We have white cube into White Cube. That will work because we can multiply two by why cute and get to y Cube. So we just add whatever we're multiplying by to the top of our fraction to the top of our division. So then we have minus two. Why squared? Plus two. Why minus two. And now we need to subtract this. We get zero plus zero plus zero plus two because this is like plus zero. So zero minus and negative two is we get plus two. But everything else is canceling. Now we're done. We can no longer perform the long division. So we write this as to why plus to whatever is on the top, plus the remainder, which is to divided by everything that's on the bottom. Everything that's on the bottom of the original fraction. So this piece right here, whatever you're dividing, dividing by. So now what this tells us is that the original integral is equal to the integral of two y plus two plus two over why cubed minus y squared. Plus why minus one because this fraction is actually equal to this to I will integrate nicely to will integrate nicely. But this piece right here is another rational function that you were most likely going to have to use partial fractions on. This was just another example of how to set up the integral how to simplify an integral using long division. The final two examples that we do in long division. We actually work through the entire in a rule. So again, this is just showing you how we work through long division and set it up. Next example we're gonna follow all the way through with solving the intervals is that the powers are matching matching powers. Is another good indication that we do need to use long division as opposed to partial fraction. So we'll go ahead and get started on the long division here. So we're gonna have nine x cubed minus three x plus one that's being divided by we draw that a little bit nicer that is being divided by X cubed minus X squared. So we need a multiplied by nine, which gives us nine x cubed minus nine x squared. And now we're going to subtract. So we get zero. We get a plus nine x squared. We get a negative three x minus zero. So we get negative three x, and we also bring down that one. There's no other way X cubed can be multiplied by anything to go into what we have left. So this becomes our remainder. So what we have is that the integral of nine x cubed minus three X plus one divided by X cubed minus X squared. Once we do the division, this is actually equal to the integral of nine. What's on top? Plus the remainder, which is nine x squared, minus three x plus one divided by X cubed minus X squared. Now nine. That looks like a nice integral but this integral is another rational function. To solve this, we're going to need to use partial fractions. So now where you're going to keep this in mind, write this down somewhere on the next page, I'm going to rewrite this integral but separately so we can solve it using partial fractions. So we have the integral of nine x squared, minus three x plus one divided by X cubed minus X squared so we can factor out an X we're left with. We can factor out an X squared, actually, and we're left with X minus one. These air both reduced the most that they can be. This is going to be equal to a over X squared plus B over X minus one. But don't forget that you also have to represent a single X because X is squared, you have to represent the squared power and the first power. Once you have this, find a common denominator. We have a X minus a plus B X squared plus C x squared times X minus one and actually this is not. This is the mistake that's commonly made B a X squared. It is already being multiplied by an X the X squared value. So it only needs to be multiplied by X and by a X minus. One on Lee needs to be multiplied by X squared. So this is good still so far. But then see is already over an ex, so it only needs to be multiplied by its Onley missing a single power of X in addition to the X minus. One factor this is going to be all over that X squared X minus one common denominator. Further factoring this, we get a X minus a plus V x squared plus C x squared minus C X all over X squared X minus one and now we can switch to just focusing on the numerator. So we're going to have that X minus a plus B X squared plus C x squared minus C X is equal to our original numerator nine X squared minus three X plus one grouping terms. We have an A X and we have a C X. We have a negative A, which is just a constant, and we have a B X squared plus a C X squared are single power of excess. We have a negative three x on the right hand side, the over one as the coefficient, and we have a nine x squared. That means on the next page we can write the equations that we're going to need to solve. So what? We can probably just do it right here. We have that a minus c is equal to negative three negative A is equal to one and B plus C is equal to nine So if negative a equals one, this implies that a is equal to negative one. If a is equal to negative one, this a first equation becomes negative. One minus C is equal to negative three, which means we end up with I see equal to two. If she is equal to two, that implies that B is equal to seven. So now on the next page, we have that A is equal to negative one B is equal to seven and C is equal to two are partial fractions were a over X squared plus B over X minus one plus C over X. So when we integrate now, we have the integral of negative one over X squared plus seven over X minus one plus two over X. And each of these can have its own integral because in a girls go through pluses. So we have negative integral of one over X squared, plus seven times the integral of one over X minus one plus two times the integral of so. This is like writing X to the negative, too. So we have negative times a negative one over X plus seven Natural log of X minus one plus and this is looks like this. Plus two natural log of X, which gives us one over X plus seven natural log of X minus one plus two natural log of X plus seat. However, this is not our final answer because this was only the second half of this was the remainder integral that we had. We also had the integral of nine. Plus all of this nine integrates to nine X. So you do have to include that. So this is what the final answer looks like being a lot of times you will have to use partial fractions on the remainder term that you come up with after doing long division. As with the situation that happened here, I started divided by X word minus two X minus three. The degree on the top is more than degree on the bottom, which means we're gonna have to use Long Division to rewrite this integral first. So we're going to use the long division. We have X squared minus two X minus three, which is going to divide two x cubed minus four X squared minus X minus three. So we get a multiplied by two X, which gives us two X cubed minus 24 minus for X squared minus 36 X subtraction. Always use your parentheses so you don't forget to distribute the negative. So two X cubed minus two x cubed zero negative four X squared plus four X squared is also zero We have minus X plus six X, which gives us a five X and then we're just gonna bring down that minus three. So what this means is that this fraction inside are integral. We can now rewrite as two x plus our remainder term, which is five X minus three. Rewrite our remainder term divided by X word minus two X minus three What we're dividing by. Okay, so two extra integrate just fine, but this five x minus three X squared minus two X minus three is going to take partial fractions. So on the next page, we're going to start that. So keep this integral in mind. But now we need to look at five x minus three, divided by X squared minus two X minus three. So, first we need to factor the bottom, which does factor nicely into X plus one x minus three So when we set up our partial fractions, we have a over X plus one plus B over X minus three. We can find a common denominator by multiplying here, and this is all over our common denominator, and then we can distribute. So now this would be like Step four of the partial fractions. We would have a X minus three a plus b x Crosby, and we would set this equal to the top of our original corruption, which is five X minus three. This means we have an A X plus a B X, which is equal to a five X, so I'm just ignoring the excess. We also have a negative three plus B, which are coefficient terms which match up with the negative three. So now we have a plus. B holds five and negative three a plus B. Who's negative? Three. We can solve the first equation and say that B is equal to five minus a. Then we can rewrite the second equation as negative. Three a plus five minus a, which is still equal to negative three. And we have that negative for a is equal to negative eight, which means a is equal to two and with a equals to be should be equal to three. So now we have that Let's go to new. So now we have that A was equal to two. So we have to over X plus one requests and be was equal to three over X minus three. And this is now what we're going to be integrating. And this is in addition to the original term we had. So let's we had the inner rule. Let me just rewrite this here to give you guys a better idea of what's going on. So we do have that A is equal to two. B is equal to three, and we also have that we have the integral of two x plus the integral of five x. So this is what we had after we did the long Division X squared minus two X minus three. And what we said was this right here. So we still have the integral of two x plus. This integral, though, is now what we use the partial fractions with, and we have the partial fractions of A, which is to over X plus one plus B, which is three over X minus three. Now all three of these in a girl's can be done. So two X integrates two X squared plus two over X plus. One is two times the natural log of X plus one and the integral. If you want to add that in here is three natural log X minus three, and this is going to be plus C. So this is another example of having to use long division and then having to use partial fractions on the remainder portion of that.

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