 In this video, we'll work through another example applying the method of partial fractions to integrate a rational function this time where a linear factor repeats itself in the complete factorization of the denominator. Be sure to follow along on your own paper as you work through the video. Consider the integral of 2x plus 4 divided by x cubed minus 2x squared. 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 the integrand 2x plus 4 over x cubed minus 2x squared. We notice that this denominator can be written as x squared times x minus 2. So we're going to set up an equation that allows us to decompose this integrand. We're going to represent the factor of x on its own with a over x. But because x is a linear factor that repeats itself, we also need a term b divided by x squared. If this were x cubed, we would then have a third term that would be plus c over x cubed. We don't, so we will move on to the next linear factor, which is x minus 2. So c divided by x minus 2. And our goal is to solve for the constants a, b, and c. We'll multiply both sides of the equation by the original denominator since it's the common denominator among all rational expressions. So if we do that, I'm going to multiply both sides by x squared times x minus 2. I end up with on the left 2x plus 4 equals a over x times x squared times x minus 2 plus b over x squared times x squared times x minus 2 plus c over x minus 2 times x squared times x minus 2. We'll rewrite that right here. And when we clean this up, we see we have ax times x minus 2 plus b times x minus 2 plus cx squared. I'm going to go ahead and distribute, expand all the terms here, ax squared minus 2ax plus bx minus 2b plus cx squared. And then I'm going to collect like terms. And I'm doing this for a reason. Let's do this first. I have a plus cx squared plus negative 2a plus bx minus 2b. Now in this case, we're going to use a little bit of a different method than we did in our previous example. We'll start it as we did before. Let's let x equals zero. If x equals zero, this tells us that 4 on the left is equal to negative 2b. So what we've done is chosen a strategic value for x that would allow us to identify one of the three letters. This then tells us that b is equal to negative 2. We're going to hold on to that. Now for a and c, we'll employ the use of systems of equations. The equal sign in our equation implies that the coefficients of each term on the respective sides of the equation are equal. So if we're equating coefficients, we find that 4 must equal negative 2b, which we just found. 2 is the coefficient of x on the left, and negative 2a plus b is the coefficient of x on the right. So we're going to set up an equation 2 equals negative 2a plus b. And then we see that there's an x squared term on the right and no x squared term on the left, which means then that zero is equal to a plus c. So I'm going to rewrite that here. So we have zero equals a plus c, negative 2a plus b equals 2, and from before we know that b is equal to negative 2. Well with b being negative 2, we can use that in this equation, negative 2a minus 2 equals 2. So negative 2a equals 4, which tells us that a is equal to negative 2 as well. And with a being negative 2, that tells us that then c must equal 2. As a result, we know then that 2x plus 4 over x squared times x minus 2 is equal to negative 2 over x plus negative 2 over x squared plus 2 over x minus 2. Again that's negative 2 in for a, negative 2 in for b, and 2 for c. Integrating this, we get the integral of negative 2x plus negative 2 over x squared plus 2 over x minus 2 dx. We're going to split this integral up. In evaluating, we get negative 2 times natural log of absolute value of x minus negative 2 over x plus 2 times the natural log of x minus 2 plus c. Using log laws, we can consolidate the natural log terms, which gives us 2 times the natural log of absolute value of x minus 2 over x. We know that this coefficient is negative 2 plus 2 over x plus our constant c.