 So, another way of doing integration through algebra is known as partial fractions. We'll need to define some terms. Let f of x and g of x be polynomials. The rational function f of x over g of x is a proper rational function if the degree of g is greater than the degree of f. It is improper otherwise. And one useful idea is that if we do have an improper rational function, we can use polynomial division to transform it into a polynomial plus a proper rational function. This leads to the following. Suppose I have two or more rational expressions. We can use the standard rules of algebra to add them to produce a single rational expression. But equality works in both directions, and what this means is that we can take a single rational expression and write it as a sum of rational expressions. This leads to the following result. Suppose I have a proper rational function where the denominator is expressible as a product of two polynomials. Then we can always produce a partial fraction decomposition where the degree of each numerator is less than the degree of the denominator. So let's try to find a partial fraction decomposition. Since we're starting with a proper rational function, our theorem guarantees we can write this as a sum of proper rational functions where the denominators are the individual factors. Here because both denominators are degree one, this means that the numerators must be of degree less than one. They have to be constants, and so we'll call them a and b. We'll multiply both sides by the common denominator x minus three times x plus two, and simplify down to a final expression. Now it's important to understand that the equals in this case means that we want the rational function to be absolutely completely 100% interchangeable with the sum of rational functions. That means that five must equal a plus bx plus two a minus three b for all values of x. And that means that the coefficients of the variables on both sides must be the same. So over on the right hand side, the coefficient of x is a plus b. So I need to make that equal to the coefficient of x on the left hand side, which is zero, because over on the left hand side there is no x, and so the coefficient must be zero. Similarly, on the right I have my constant term two a minus three b. This must be equal to the constant term on the left, five. And now this gives me a system of two equations with two unknowns a and b, and I can solve that system of equations, and I have a equals one, b equals negative one. And again, this is the correct way to solve for a and b. However, the usual way of solving this system is slightly different. And this is based on the idea that we need this equation to be true for all values of x, or since this equation came from the preceding equation, we need this equation to be true for all values of x. So what's a good value of x? Well notice that if x equals negative two, this first term drops out, and so we get the equation, which is an equation that's much easier to solve for one of the variables. How do we get the other variable? We'll choose a different value of x, and a convenient value of x to choose is going to be x equals three, because that's going to make this second term drop out. And if I let x equals three, we get the equation, which is again much easier to solve for our variables, and we have the same solution a equals one, b equals negative one. Now before proceeding, we might want to think about why we can't actually do it that way, but why we still do it that way. So remember that our original equation was, and in this equation, x can never be three or negative two. And so when we dropped x equals three, x equals negative two, and produced an equation in a and b, we were taking advantage of a situation that could never exist. So why did it work? Part of the reason that it worked is that any value of a and b that satisfies this equation for all x other than three or negative two will also satisfy this equation for all x other than three or negative two. So again, all we really need to do is to find an a and b that satisfies this equation for all x other than three or negative two. Well, we did one better. We found an a and b that satisfies this equation for all x, which means it will certainly satisfy our original equation for all x other than three or negative two. It's important to understand that everything up to this point is just algebra. We've done no calculus so far. So let's go ahead and introduce a little bit of calculus and talk about the anti-derivative of five over x minus three times x plus two. Because the denominator is in factored form, we can find the partial fraction decomposition. Actually, we've already done that. And the additivity of the anti-derivative means we can split this integrand into two integrands and evaluate each one separately. So we'll use a u-substitution. For the first integral, we'll let u equals x minus three, so du is dx, making our substitutions, integrating, and we'll put everything back where we found it. And to take into account the fact that you can't take the log of a negative number, we'll throw the argument x minus three into a set of absolute value bars. Similarly, for the second integral, we'll let u equals x plus two and then du will be dx. We'll make our life easier by factoring that negative one out front of the integral. We'll make our substitutions. We'll find the anti-derivative. We'll put everything back where we found it. And again, because this is log, we'll want to take the log of the absolute value of x plus two.