## This problem is FRUSTRATING me SO Much!

1. The problem statement, all variables and given/known data
Find the indefinite integral...
X^2/ X-1

In my book they set it up as 2 separate integrals...
X+1 + 1/X-1

I figured they got that from long division but I am not getting that answer. So can someone please tell me how they got those 2 separate integrals
 $$\int\frac{x^2}{x-1}dx$$ $$\int\frac{x^2-1+1}{x-1}dx$$ $$\int\left[\frac{(x+1)(x-1)}{x-1}+\frac{1}{x-1}\right]dx$$ $$\int\left(x+1+\frac{1}{x-1}\right)dx$$ Does this make a little more sense?
 Yes it makes more sense but I just want to make sure I understand what you did.. In part B you completed the square...(But I thought it would be (X^2 + 1) - 1)

## This problem is FRUSTRATING me SO Much!

I did not complete the square, and plus, you do not have a linear factor - so you shouldn't even consider it.

$$x^2-1=(x+1)(x-1)=\mbox{Difference of Squares}$$
 Oh....I don't think we've used that before...I will have to look it up...thank you!

 Quote by BuBbLeS01 Oh....I don't think we've used that before...I will have to look it up...thank you!
BuBbLeS, you have used it plenty of times (you learn it in junior high or HS). You just forgot! No worries, go look it up and you'll be like oh yeah ... I remember that :p
 I hope so lol! I have another question on another problem... I am trying to complete the square... I have 6x - x^2 Then I did (6/2)^2 = 9 So I end up with (-x^2 + 6x + 9) - 9 I need to end up with 9 - (x - 3)^2 Can I make (-x^2 + 6x + 9) to (x^2 + 6x + 9) because the x is squared so it will be positive or is that negative mean like -(x^2)?? And (x - 3)^2 gets me x^2 - 6x + 9 so I am a little confused...
 Before you complete the square, make sure that the coefficient of your leading term is a positive one. $$6x-x^2$$ $$-(x^2-6x+9-9)$$ $$-(x^2-6x+9)+9$$ $$9-(x-3)^2$$
 Ohhhh....thank you so much!!!
 So you solved it, yes?