 In this video, we'll work through an example applying the method of partial fractions to integrate a rational function. Be sure to follow along on your own paper as you work through the video. Consider the integral of 1 over x squared plus x minus 2. Our goal is to decompose the integrand into partial fractions so that we can use known methods of integration to evaluate the integral. Let's focus on this integrand 1 over x squared plus x minus 2. Now we notice that the denominator can be rewritten as x minus 1 times x plus 2. So we set up an equation to find an equivalent expression for 1 over x squared plus x minus 2 using the factors in the denominator. 1 over x squared plus x minus 2 equals some constant A over x minus 1, one of the factors, plus B, some other constant, over x plus 2, which is the other linear factor of the denominator. Our goal is to solve for the constants A and B. So let's do this by multiplying both sides of the equation by the original denominator, x squared plus x minus 2, since it's the common denominator among all rational expressions. If I multiply both sides by the original denominator, I obtain the following. 1 equals A times, now since this denominator can also be written as x minus 1 times x plus 2, we see that when we distribute it with this term A over x minus 1, we're left with x plus 2 in the numerator. Plus B times x squared plus x minus 2 divided by x plus 2 leaves me with x minus 1 in the numerator. So we're going to employ a sort of a trick to find A and B separately. So I'm going to rewrite that previous step. Let's suppose that x is equal to negative 2. What we know about this expression is that it's supposed to be true for any x. So I have freedom to choose whatever x value I want to help me get to my ultimate goal. So if we let x equal negative 2, what that allows us to do is drop this term, because when x is negative 2, we have A times 0. So when x is negative 2, I have 1 equals A times 0 plus B times negative 2 minus 1. Simplifying, we have 1 equals then negative 3B, which then tells me that B is negative 1 third. I'm going to hold on to that. I'm going to do a similar thing with supposing that x equals 1. Notice I'm choosing x values strategically. Those that will allow me to drop one of the terms in this equation. When x is 1, I have the equation 1 equals A times 1 plus 2 plus B times 1 minus 1. This tells me that 1 is equal to 3A, which then tells me that A is equal to 1 third. So we've now found that 1 over x squared plus x minus 2 is equal to 1 third over x minus 1 minus 1 third over x plus 2. So our original integral can be rewritten as you see here, the integral of 1 third over x minus 1 minus 1 third over x plus 2. We basically decompose the original rational function into two rational functions that we know how to integrate. This is equal to the integral of 1 third over x minus 1 minus the integral of 1 third over x plus 2. We know this as 1 third times the natural log of absolute value of x minus 1 minus 1 third times the natural log of x plus 2 plus C since this is an indefinite integral. We can then use log laws to clean this up by writing 1 third times natural log of the absolute value of x minus 1 over x plus 2 plus C.