
#1
Mar1608, 10:50 AM

P: 605

1. The problem statement, all variables and given/known data
Find the indefinite integral... X^2/ X1 In my book they set it up as 2 separate integrals... X+1 + 1/X1 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 



#2
Mar1608, 10:52 AM

P: 1,757

[tex]\int\frac{x^2}{x1}dx[/tex]
[tex]\int\frac{x^21+1}{x1}dx[/tex] [tex]\int\left[\frac{(x+1)(x1)}{x1}+\frac{1}{x1}\right]dx[/tex] [tex]\int\left(x+1+\frac{1}{x1}\right)dx[/tex] Does this make a little more sense? 



#3
Mar1608, 10:59 AM

P: 605

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) 



#4
Mar1608, 11:00 AM

P: 1,757

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.
[tex]x^21=(x+1)(x1)=\mbox{Difference of Squares}[/tex] 



#5
Mar1608, 11:04 AM

P: 605

Oh....I don't think we've used that before...I will have to look it up...thank you!




#6
Mar1608, 11:07 AM

P: 1,757





#7
Mar1608, 11:25 AM

P: 605

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... 



#8
Mar1608, 03:38 PM

P: 1,757

Before you complete the square, make sure that the coefficient of your leading term is a positive one.
[tex]6xx^2[/tex] [tex](x^26x+99)[/tex] [tex](x^26x+9)+9[/tex] [tex]9(x3)^2[/tex] 



#9
Mar1708, 08:14 PM

P: 605

Ohhhh....thank you so much!!!



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