 Let's look at a few more cases involving the integration of rational functions. For example, consider the integral 3x squared plus 4x plus 2 over the product x plus 3 times x squared minus x plus 5. Now the first thing to remember is that you can only apply the rational function decomposition theorem to proper rational functions. So we should verify that the numerator is a polynomial of degree 2. The denominator is a product of a degree 1 and a degree 2 polynomial, and so the denominator is a polynomial of degree. So this is a proper rational function. Since the denominator is already factored, our theorem guarantees we can rewrite it as a sum of two rational expressions. Where the denominators are the individual factors x plus 3 and x squared minus x plus 5, and the numerators are expressions with a degree 1 less than the denominator. So the numerator of x plus 3, a first degree polynomial, is a constant, which we'll call a, and the numerator of x squared minus x plus 5, which is a second degree polynomial, is some first degree polynomial, which we'll call bx plus c. Now since everybody loves working with fractions, we'll go ahead and find a common denominator and add, wait, wrong script. What I meant is that we really don't want to work with the fraction, so let's multiply through by that common denominator, x plus 3 times x squared minus x plus 5. And after all the dust settles, we have a much nicer expression to deal with, and we'd like to make this statement true for all values of x. So we'll pick a value of x that we want to work with, like pi over square root of 17. In this case, we might consider this factor. If x equals negative 3, this factor is 0, and so the entire term is going to drop out, and we won't have to deal with it, and we'll have a much simpler equation. So let's make sure it's true for x equals negative 3. If x equals negative 3, solving this equation, then we find that a must be equal to 1. Now remember that we chose x equals negative 3 because that would make one entire term disappear. However, we have a problem. The x value that will make this first term disappear is going to be an irrational number, and we don't really want to work with a complicated expression. Fortunately, making one of our terms disappear isn't required, it's just something that happens to be useful on some occasions. If we can't do that, or if it requires using some irrational value for x, we could always choose any value of x we want to work with. For example, if x equals 0, solving this equation, we find that c is equal to negative 1, and if x equals 1, solving this equation, and we find that b is equal to 2. It's important to keep in mind that all of this is just algebra. We haven't done any calculus at all, so let's start doing the calculus. First, because our integral is a sum, we can split it. Let's do a u-substitution on that first integral, so we'll let u equals x plus 3, and since it's possible that x plus 3 might be a negative number, we do need to throw the whole thing in absolute value. The second integral will let u equals x squared minus x plus 5, and so d u will be, and luckily enough, that's exactly what our numerator is, so we can do a u-substitution. And again, we'll get the log of x squared minus x plus 5, which will enclose in an absolute value operator. Bonus points for those of you who recognize we don't actually need the absolute value operator in this particular case, and we get our final expression.