integrate (x + 1)^3 / (x - 1)(x +2) dx
A little reminder here. A fraction of the form (mx + c)/ (x+2)(x-1) is seperaple into the form a/(x+2) + b/(x-1) which is easy to integrate. But if the degree of the numerator is greater then that of the denominator, partial fractions cannot be used.
In this case, the degree of the numberator, is greater then the degree of the denominator, so we cannot use partial fractions....yet.
These problems usually involve some rearanging and seperating before you can solve it. Also adding +u and -u to the expression to rearrange it into a desired form.
Well I worked with this problem for like 2 hours and eventually, by looking at the answer, determined how the book solved it.
∫ (x + 1)^3 / (x - 1)(x +2)
add + (x+2)(x+2)(x-1) and - (x+2)(x+2)(x-1) to the numerator (0)
(x + 1)^3 + (x+2)(x+2)(x-1) - (x+2)(x+2)(x-1) / (x - 1)(x +2)
this can be separated as the sum of two fractions:
(x + 1)^3 - (x+2)(x+2)(x-1) /(x - 1)(x +2) + (x+2)(x+2)(x-1) / (x - 1)(x +2)
the second fraction = x + 2 and is a piece of cake to integrate, if you multiply out the first fraction and add like terms, it ends up being like 3x + 5/ (x-1)(x+2), in this the degree of the numerator is less then the degree of the denominator, so we can now use partial fractions to solve. And this produced the correct answer.
Ok, it might have taken me years to guess that! What I'm wondering, is there some sort of clue they gave me in the way the problem is set up? Are there some clues I missed to help me guess that earlier?
These are starting to come up more frequently and I'm not sure if I'm supposed to be just guessing like this. I know integration involves a lot of guessing, trial and error, but this one is crazy!
A logarithm is just a misspelled algorithm.
Nope, I'd have done it in exactly the same way as you. These questions tend to be very tedious, but they also aren't too hard, just time-consuming, and they give you a lot of marks in exams because of the time you spend doing them.
Why did the vector cross the road?
It wanted to be normal.