 This lesson is on the method of partial fractions. You may have done partial fractions before in an algebra class, but now we expand it to our integration in calculus. So let's go on. What is a partial fraction? It is a re-expressing, a rational function, a ratio of polynomial functions if you remember from algebra, as a sum of simpler fractions. Let's do an example on this re-expressing. Suppose we have the problem x minus 3 over x squared minus 4, and we want to create simpler fractions. Many of you have just done these in your head before, but there is this method called partial fractions. What we do is set up the fraction x minus 3 over x squared minus 4, and let that equal to a over a factoring of the x squared minus 4. Well, this would be x minus 2 times x plus 2, so we will put one of the factors here. And of course, if we have a over x minus 2, we need something else, which is b over x plus 2. We have the lowest common denominator here when we finish doing our re-expressing. So now we have to figure out what a and b are, and this is the process we use. We do x minus 3 is equal to put the lowest common denominator of x minus 2 times x plus 2, which means x plus 2 multiplies a, so we have a times x plus 2 plus the b of x minus 2. We substitute in whatever makes this factor 0, and then we substitute it in whatever makes that factor 0. So let's make x equal to negative 2. If we put a negative 2 into the x on the left-hand side, we get negative 5, and that's going to be equal to, this term becomes 0, this becomes negative 4b. Therefore, we can solve for b and make it equal to 5 fourths. Then we substitute in x is equal to 2. Once we do that, the b part cancels out, and we're left with the a. x equals 2, substituted in on the left-hand side, gives us negative 1 is equal to 2 and 2 is 4, 4a. So now a is equal to negative 1 fourth. So now we have re-expressed our x minus 3 over x squared minus 4 to be equal to a, which is negative 1 over 4 times x minus 2, plus rb, which is 5 over 4 times x plus 2. So let's go on and work this into a problem of integration. Now let's integrate this. We have the problem, integrate x minus 3 over x squared minus 4 dx. We can set this up into our partial fractions, which means we want the integral of negative 1 fourth times 1 over x minus 2, and you'll see I'm writing this a little bit differently, plus 5 fourths times 1 over x plus 2, and of course we need rdx. So if we do that, the integration becomes quite simple. It's negative 1 fourth ln of absolute value of x minus 2 plus the 5 fourths ln of the absolute value of x plus 2 plus a constant. We can continue this problem and simplify it a little more and get negative ln of the absolute value of x minus 2 to the 1 fourth power plus ln of the absolute value of x plus 2 to the 5 fourths power plus c. That goes on to ln of the absolute value of x plus 2 to the 5 fourths over absolute value of x minus 2 to the 1 fourth, and we can make plus c. And we can make one final revision and make it equal to 1 fourth ln of the absolute value of x plus 2 to the 5th power over absolute value of x minus 2 plus a constant. You will see them written either way in the long hand form or this division form. Well, let's try another example using partial fractions. Let's integrate x minus 3 over x squared plus 4x. Let's break this up into partial fractions. Well, first of all, we need to factor our denominator. So that's x times x plus 4, and then set that up into a over x plus b over x plus 4. So now we have x minus 3 is equal to a times x plus 4 plus b times x. Setting x equal to 0, we get negative 3 is equal to 4a, or a is equal to negative 3 fourths. Setting x equal to negative 4, we get negative 7 is equal to negative 4b, and that goes to b equaling 7 fourths. Set this back into our integral. So we have a, which is negative 3 fourths. So we'll have negative 3 fourths times 1 over x plus b, which is 7 fourths, times 1 over x plus 4 dx. Integrating this, we get negative 3 fourths ln of the absolute value of x plus 7 fourths ln absolute value of x plus 4 plus our constant. Again, we can put these together if we so desire, or just leave them separated. Not too bad. This one is similar to the last one. Let's go on to another problem. In this particular problem, we have to divide first. So let's do that. And this has to be a long division. x squared minus x minus 2 divided into x squared plus 3. x squared goes into x squared one time. If you haven't practiced long division a long time, try it. So we're going to have x squared minus x minus 2. And if you remember in long division, we just change signs when we get to this point. So we will be left with x plus 5. So this integral now becomes 1 plus x plus 5 over x squared minus x minus 2 dx. Now we must separate our x plus 5 over x squared minus x minus 2 into partial fractions. So we go x plus 5. And of course, if we factor this, we get x minus 2 times x plus 1. Normally, these factorings will be very simple. It's equal to a over x minus 2 plus b over x plus 1. Now, why did we divide first? Because we could not do an anti-derivative or an integration unless we did that. Anytime the degree in the numerator is greater than or the same as the degree in a denominator of an integral, you must divide. So remember that no matter what you see, whether you go into this partial fractions or not, you will always need to divide. So let's go on with the problem. So we'll have x plus 5 is equal to a times x plus 1 plus b times x minus 2. So if x is equal to negative 1, that will leave us a 4 on the left hand side, and that is equal to negative 3b. And that gives us b to be negative 4 thirds. If we make x equal to 2, that gives us 7 is equal to, the b's will go out this time, so that's 3a. So a will be equal to 7 over 3. Setting this up again, we'll have integral of 1 plus a, which is 7 thirds, times 1 over x minus 2, plus b, which is negative 4 thirds, times 1 over x plus 1. Of course all of that is dx. And this time we have to take that integral of 1, which is x plus 7 thirds ln of the absolute value of x minus 2, minus 4 thirds ln of the absolute value of x plus 1, plus our constant. Last but not least, let's look at this problem. We want you to pause the video and try this. Determine the integral of x squared minus 2 over x cubed plus 3x squared plus 2x. This problem of course lends itself to partial fractions again. And we don't have to worry about dividing because the degree of the numerator is smaller than the degree of the denominator, but we do have to factor the denominator. And it's a double factoring. First we'll take out the x and get x squared plus 3x plus 2 and factor the trinomial and get x plus 2 times x plus 1. So now we will have x squared minus 2 is equal to a over x plus b over x plus 2 plus c over x plus 1, a triple 1. Let's clean this up a little bit. We'll have a times the x plus 2 times x plus 1, if our lowest common denominator is x times x plus 2 times x plus 1, plus bx times x plus 1 plus cx times x plus 2. Now there are many different ways to do these types of problems, but I have found the simplest method is with our substitution. It makes everything clean and neat and you'll see. If I set x equal to 0, what happens? Well the b term and the c term cancel out. I am only left with the a. So we'll have a 0 substituted for x on the left hand side and get a negative 2. And when I substitute it on the right hand side I get 2 times 1 is 2a. So that gives me a equal to negative 1. Very simple. Try the other ones. x is equal to negative 2 in this one. a will cancel out and c will cancel out. I am only left with b. So going into the left hand side, negative 2 squared is 4 minus 2, which is 2 is equal to 2b. So b is equal to 1. Next we'll get c. x is equal to negative 1. So in this case a will cancel, b will cancel. We'll just be left with the c. So put the negative 1 in x squared minus 2 and we'll get negative 1 equals negative c. So c is equal to 1. Let's create our integral. So now we have the integral of negative 1 over x plus 1 over x plus 2 plus 1 over x plus 1 dx. And we get negative ln of the absolute value of x plus ln of the absolute value of x plus 2 plus ln of the absolute value of x plus 1 plus our constant. And this ends our lesson on the method of partial fractions.