When doing this problem, we have to start with doing long division because the degree of the numerator is more than the degree of the denominator. So when we do long division, we end up with X ^3 + X + X + 24 / X ^2 - 1. Now I'm just going to look at that X + 24 / X ^2 - 1 portion because we know in the final answer we're going to have the X ^3 + X. We're trying to do the partial fraction on that X + 24 / X ^2 - 1 X squared -1. We know factors in the X + 1 X -1. So if I put a / X + 1 + b / X - 1, we cross multiply. We get a times the quantity X -, 1 + b times the quantity X + 1 equal in the X + 24, distribute it out, and then we're going to take the X coefficients, so one is going to equal A+B. Then we're going to take the terms that don't have an X in them. 24 is going to equal negative A+B. If we add those two equations exactly the way they are, we end up with 2B equal 25 or B is 25 halves. If we take that second equation and multiply through by -1, we could have actually taken either of them. We end up with -23 equaling 2A or A equal -23 halves. Now perhaps some of you are wondering, what if I put the a over the X - 1 and the B over the X + 1? It really doesn't matter because when we solve it, the A will be whatever was over the X - 1, whereas in what I just did, the B was over the X - 1. So doing the same steps, we'd have X + 24 equaling. Now this time a * X + 1 + b * X - 1 or AX plus A+ BX minus B one equal A+B 24 equal a -, b Adding them together we get a equal 25 halves. Taking one or the other equation to multiply by a -1, we get 23 equal -2 B or -23 halves equal B. So it doesn't matter where the A and the B go, because as we solve them they are over the respective terms. So our final final answer is going to be X ^3 plus IX plus 25 / 2 times the quantity X -, 1 - 23 / 2 * X + 1.