 Welcome back to our lecture series Math 1220 Calculus II for students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Misseldine. We're going to be starting in lecture 14 actually over the next two lectures, 14 and 15. We're going to start our discussion about partial fraction decompositions and how these can be useful for calculate anti-derivatives of rational functions, that is, a polynomial divided by a polynomial. Now, before we do that, I want to remind us that we're quite familiar with the process of adding fractions or other algebraic expressions. So consider the following situation. Let's say we have the fraction 3 over x plus 4. We want to add it to 2 over x minus 3. So what do we do? We have to first find the least common denominator, the LCD for short. And oftentimes, if there's no common factors between the two denominators, in this case, we have x plus 4 and x minus 3, there's no common factors. The least common denominator would be just the product of those two, x plus 4 times x minus 3. So we first find the LCD, and so then we have to rescale the fractions. So with the 3 over x plus 4, it's missing the x minus 3 in the denominator. So we have to times the top and bottom by x minus 3. That's what you see right here. And then for the second one, it's missing the x plus 4 in the denominator. So we have to times top and bottom by 3, or x plus 4 on top and bottom. That's what you see right here. So we have to rescale the fractions so that they have a common denominator. You can see now, let me get this out of the way, you can see now that the denominator here is identical to the denominator here. So therefore, we could add the numerators together. Now to add them effectively, you're going to distribute the 3 here onto the x into the negative 3. This will give us a 3x minus 9, and then distribute the 2 onto the x and the 4. This gives you a 2x plus 4. And so then adding like terms, we get 3x plus 2x, which is a 5x, and we get negative 9 plus 8, which is a negative 1, like so. Now in this situation, you'll notice I also multiply out the denominators x plus 4 times x minus 3. If you foil that out, you're going to get first x squared outside negative 3x inside 4x and last negative 12. And then in this situation, combining like terms, you get negative 3x plus 4x, which is just a positive x. The denominator would become x squared plus x minus 12. So the idea here is that all of these terms, I'm going to erase this part here as well, that the sum of these two fractions adds up to be this combined form over here. Now frankly speaking, I think it's a horrible idea to ever multiply out the denominator of a rational function. We do so much better when they're factored. And for example, identifying the domain is super easy when they're factored. You can't have x be negative 4 or 3. Here it's much more mysterious, but with that in mind, it's very common place that people multiply out the denominators. I would swear us under a blood oath never to multiply out the denominator, but like I said, if we're given a rational function, it could be the denominators already multiplied out and we might have to factor it. So we're quite used to this process here where we add two fractions and end up with something like here. It turns out for the sake of integration, it's going to be very useful to reverse this process. What I mean by that is to take the following scenario. Let's say that we want to integrate the rational function 5x minus 1 over x squared plus x minus 12. Now, if we were trying to integrate this thing, we might be tempted to do something like a u-substitution, seems like a good idea. Take u to be x squared plus x minus 12, in which case du would equal 2x plus 1 times dx. And unfortunately, no amount of substitution, sorry, no amount of manipulation is really going to get us a 2x plus 1 in the numerator here, at least not in any convenient way. In which case, we're kind of like, what are we going to do? Well, the good news is, if we know that this rational function or integrand is equal to these two partial fractions, we could then replace the big fraction with these two smaller fractions, these two partial fractions, what they're called. So this would be the same thing as the integral of 3 over x plus 4 dx, plus the integral of 2 over x minus 3 dx. Taking out the 2 and the 3 here, as they're just constant coefficients, we get 3 times the integral of dx over x plus 4, plus 2 times the integral of dx over x minus 3. And why this is a better scenario is, in this situation, what if we try the u-substitution? You take u to be x plus 4, then you take du just to be dx, well that's kind of nice. This thing would look like 3 times the integral of du over u. And hopefully, this is when we remember, the antiderivative of 1 over u is the natural log, 3 times the natural log of the absolute value of u plus the constant. And so if we use that fact right here, this first integral, the 3 over x plus 4, its antiderivative will be 3 times the natural log of the absolute value of x plus 4. You do need to have the absolute value here, because x plus 4 could be negative for different choices of x. And then for the second one, its antiderivative is going to be 2 times the natural log of x minus 3 plus a constant. And this right here does give us a pretty good antiderivative. If you wanted to rewrite it using properties or logarithms, you could do that. You could bring the 2s and the 3s inside, you could combine some terms. This could look like the natural log of the absolute value of x plus 4 cubed times x minus 3 squared. That would also be correct. I mean, I'm perfectly happy with the first one. The logs do not need to be combined together, but the idea here is that if we knew that this complicated rational function on the right could be written as these two partial fractions, then we could very easily calculate the antiderivative, but how does one reverse this process? We know how to go from the partial fractions to combined fractions, but how do you reverse that? So in section 7.4 of Stuart's textbook, we want to talk about this technique of partial fraction decomposition so that we can utilize it to help us calculate antiderivatives of rational functions. And we'll do some examples of this in the next couple of videos.