 Let's evaluate the integral of x cubed plus six over x minus one with respect to x here. Now this is a rational function, x cubed plus x over x minus one. So in order to find the antiderivative, what we wanna do is use the technique of partial fraction decomposition to rewrite this rational function as a sum of simpler partial fractions. Now the first step in a partial fraction decomposition is to check whether the rational function is proper or improper. And so this is by analog what we mean with integer fractions, right? If you took the fraction, for example, seven halves, this is typically what one means by an improper fraction. The numerator, seven is larger than the denominator. So it's an improper fraction. We could write this as a mixed number, right? How does one do that? Well, if you're not sure what to do, you always turn this into a division problem, right? Seven over two. How many times does two divide into seven? Well, six is a multiple there, you get three. So track them, you get one, that's your remainder. So seven divided by two is three, minor one. What that means is you get three and one half if you write this as a mixed number. Which of course three and one half, the notation is kind of weird because you put the things side by side, it almost seems like it's multiplication. Really it's addition, it's three plus one half. One can do a similar thing for rational functions as well. So this right here is an example of an improper rational function. The numerator is bigger than the denominator. And what I mean by that is the exponents, right? The numerator is a degree three polynomial. The denominator is degree one. Degree three is bigger than one. So what we wanna do is we wanna write this as a mixed number or the equivalent of that as we want some type of mixed polynomial. So we wanna write this with a whole number part, which the whole number part in this case will be a polynomial. And then over here, the fraction part is gonna be a rational function, but it's gonna be a proper rational function where the numerator is smaller than the denominator. And how do we accomplish this? Well, we have to do long division of polynomials. Let's remind ourselves how we do that. We're gonna take x cubed plus x and divided by x minus one, like so. So analyzing the leading terms, we have this x cubed and this x. We have to ask ourselves, how many times is x divided to x cubed? Our thinking of the ratio x cubed divided by x. That supplies just to be an x squared. And that's what we're gonna record on the top of our quotient bar right here. So x divides into x cubed, x squared times. So we record that on the top. The next thing we're gonna do is we're gonna times the x minus one by x squared. So x squared times x minus one, that gives us x cubed minus x squared. And so we're gonna record that over here. x cubed minus x squared. And now we have to subtract. This is mimicking exactly what we did over here, right? We look for the biggest multiple of two that goes into seven. That was two times three, which was six. We then subtract that from the dividend. And so we see here, if you take x cubed minus x cubed, those will cancel out. The leading term should always cancel out here. You're gonna take zero x squared minus minus x squared. Make sure this negative sign distributes onto the whole thing. So that's actually gonna be a positive x squared. And then bring down the next term, which we get an x. In this situation, there was no multiple digits, but if we had multi-digits over here, you would bring the term down and repeat the process. So now we ask ourselves, how many times does x divide into x squared? That happens x times. And so we record that on the top plus x right there. Next, we have to take x times it by our divisor. And so we get x times x minus one. That's gonna be x squared minus x. And we record that below here, x squared minus x. Then we subtract it. Make sure you subtract it from the whole thing right here. x squared minus x squared, the leading term, if you chose it correctly, should always cancel out. Then we're gonna get an x minus negative x. So that's actually a two x. There's no terms to bring down, which really just means you're gonna bring in a zero right here. Give myself a little bit of space. And we repeat this process one more time, all right? Two x divided by x is gonna be a two. So we write that on the very top, that's our quotient. Two, we then have to take two times x minus one. Two times x minus one, that gives us two x minus two. Bring that down here, two x minus two. Subtract it, make sure you subtract the whole thing. The two x's should cancel. And you're left with zero minus a negative two, which is gonna be a positive two right here. And this is gonna be our remainder term. So it took us a while here, but x cubed plus x divided by x minus one turned out to be this thing right here. x squared plus x plus two with a remainder two. So what does that mean for us? Let's see how much space I hear at the top. I'll erase some of these things right here. So what we now can see is that, well, actually, I'll just have to bring it down. It's too crowded there. So what we've seen is the following. We saw that through the long division we just did there, the, and I'll bring it so it's on the screen again. So we can see what we're doing here. So we've seen that the rational function x cubed plus x over x minus one, this will then become, you have the whole number part, which is the quotient x squared plus x plus two. Think of this as a mixed number, you have the whole number part, then you have a proper fraction. Well, that fractions will have the same denominator we started off with, which is x minus one, but now the numerator will be the remainder, which is two. So x cubed plus x divided by x minus one is the same thing as x squared plus x plus two plus two over x minus one. And so in terms of integrals, if we integrate this with respect to x, we integrate this with respect to x, we can then break up the second part into its pieces. So we have the integral of a polynomial, integral antiderivative polynomials are pretty nice, pretty simple. So we're gonna love that. But then look at the rational function we have here. You have two over x minus one. I'm gonna take the two out and we're left with this x minus one. This one does the basic U substitution saves the day. Take U to be x minus one, take du to be just dx. And then therefore this integral has the form the integral of du over u. Its antiderivative is the natural log of u plus a constant. Don't forget the absolute value as well. And so applying this to the whole thing, we take the antiderivative of the polynomial, we get x cubed over three plus x squared over two plus two x. And then we're gonna end up with, for the second part, two times the natural log of x minus one, absolute value of x minus one. Don't forget the absolute value plus a constant right there. And so we can find an antiderivative when we take the long division of the polynomials in this situation here. And so the advice I'll give you is that whenever you have a improper rational function, find the quote unquote mixed polynomial, find the quotient and the remainder using long division. Now in this situation, since the denominator was x minus one, you could have gotten away with a simpler technique called synthetic division, which I'm not gonna demonstrate right here, but that applies in some special situations. Most likely you'll have to use long division, which is why I wanted to use it here. So with the improper rational function, use long division to find the quotient and the remainder, and then integrate the quotient and remainder separately and now give you the antiderivative there. Now the issue is that the remainder is gonna be a proper fraction. And so this situation wasn't so complicated, but we'll see in the next videos, what do you do when the proper fraction's a little bit more complicated than just having a linear denominator? Stay tuned.