1 6 = x 2 - 2 x + 2 ln x - - 2 ln x - 3 + c x 2 In the previous example there were actually two different ways of dealing with the x2 in the...

1 6 = x 2 - 2 x + 2 ln x - - 2 ln x - 3 + c x 2 In the previous example there were actually two different ways of dealing with the x2 in the denominator. Which gives the following two terms in the decomposition, A B + x x2 We used the second way of thinking about it in our example. Notice however that the two will give identical partial fraction decompositions. So, why talk about this? Simple. This will work for x2, but what about x3 or x4?