 There's one more situation we can handle using a reduction by partial fractions. What happens if we have a repeated factor? In that case, we can use an extension of our partial fraction decomposition. So suppose my rational function looks like a polynomial over the nth power of some polynomial, and further suppose it's a proper rational function. Then we can always produce a partial fraction decomposition, where our denominators are going to be the powers of our repeated factor, and each numerator is a polynomial whose degree is less than the degree of our repeated factor. So let's take a look at that. Let's say I want to find the anti-derivative of x over x plus 3 squared. So notice that in this case our denominator is the power of a polynomial. It's x plus 3 squared. So the theorem says that I can write this as a sum of proper rational functions, where the denominators are the powers of x plus 3, and the numerators all have degree less than the degree of x plus 3. Since x plus 3 is a first degree polynomial, that means our numerators must be 0th degree polynomials. They must be constant, in other words. So I'll multiply through by x plus 3 squared, and I'll pick some convenient values for x. For example, if x is negative 3, then the x plus 3 term will drop out, and I'll have the equation. And I need a second value for x to get a second equation, so maybe I'll try something easy like x equal to 0, which will give me the equation. And now I have a system of two equations in two unknowns, and I can solve for a. So that tells me the numerators of my partial fraction decomposition. I can use the additivity of the integral, and now we have to evaluate these integrals separately. So let's use a u substitution. So I'll let u equals x plus 3, then du is dx, and we can make that substitution, integrate, put things back where you found them, including the absolute value around the argument. As for the other integral, we can use the same u substitution, but let's remove that common factor to outside the integral before making the substitution. We'll integrate, and put things back where we found them. The main difference between a hard problem and an easy problem is the number of steps. Most hard problems are a lot of easy problems strung together. So because this is a proper rational function, we might try the method of partial fractions, and so our partial fraction decomposition theorem guarantees we can split this into the sum of two fractions with the factors being the denominators. Since we have a repeated factor here, we can split this further into two fractions where our denominators will be the powers of the repeated factor, along with the other term, and all of the numerators will be of a degree 1 less than the individual factors, so they should all be of degree 1. Multiplying by our common denominator, and that'll give us our equation, which we want to be true for all values of x. Since the equation is supposed to be true for all values of x, it has to be true for any value of x that we choose, and so a convenient value of x to choose would be x equals negative 1, which will make these two terms drop out and leave us with only the term with the unknown b. So we'll substitute, and solve, giving us b equals 5. Another convenient value of x to choose will be x equals 3, which will make these first two terms drop out, leaving us only with the term containing the variable c, we'll substitute, and solve, giving us c equals 2. Now there's no other value of x that will make two of our three terms drop out, but we can pick any value of x that we want. So let's choose about x equals 0. We'll substitute, then solve our equation, which gives us a equal to 1. And so our integrand can be rewritten as a sum of rational functions, and additivity property of the integral allows us to rewrite this as, and evaluate each integral separately, gives us our final answer.