Why do we choose to multiply by 2 in simplifying this equation?

Click For Summary

Homework Help Overview

The discussion revolves around the simplification of an integral involving a rational function. Participants are examining the steps taken to rewrite the integrand for easier integration, particularly focusing on the choice of multiplying by 2 in the numerator and denominator.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the rationale behind multiplying the numerator and denominator by 2, questioning why this specific number is chosen over others. They discuss how this manipulation aids in separating terms for integration.

Discussion Status

Some participants express understanding of the simplification process, while others seek clarification on the reasoning behind specific choices made in the manipulation of the equation. There is an ongoing exploration of alternative methods and their effectiveness.

Contextual Notes

Participants are working within the constraints of a homework problem, which may limit the information available for discussion. The focus is on understanding the algebraic manipulation rather than arriving at a final solution.

trajan22
Messages
134
Reaction score
1
the main problem I am having is i can't figure out how they simplified this equation. Ill try and use latex but its my first time so i don't know if it will work.
[tex] \int(1-\frac{1}{3(x+1)}+\frac{(x-2)}{3(x^2-x+1)}) dx[/tex]
and somehow this simplifies to this
[tex] x- (1/3)ln(abs(x+1)) +(1/6) \int \frac{(2x-1)}{(x^2-x+1)} dx - (1/2) \int \frac{(dx)}{(x-.5)^2+(3/4)} [/tex]
i understand how to get the x-(1/3)ln(2x-1) but i don't get how they separated the other two terms for integration. or how they pulled out (1/6)
 
Last edited:
Physics news on Phys.org
Given [tex] \int\frac{(x-2)}{3(x^2-x+1)}\, dx[/tex], multiply the numerator and denominator by 2 and rewrite the numerator as (2x-1)-3.
 
Ok, well the third term in the integrand of the original question is [tex] \frac{(x-2)}{3(x^2-x+1)}=\frac{(2x-4)}{6(x^2-x+1)}=\frac{(2x-1)}{6(x^2-x+1)}-\frac{3}{6(x^2-x+1)}=\frac{1}{6}\frac{(2x-1)}{(x^2-x+1)}-\frac{1}{2}\frac{1}{(x^2-x+1)}=\frac{1}{6}\frac{(2x-1)}{(x^2-x+1)}-\frac{1}{2}\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}[/tex] where the last step in the denominator of the second term is done by completing the square.
 
ok i understand what is going on now, but why do we chose to multiply the numerator and denominator by 2, and why do we chose 2 rather than any other number.
 
trajan22 said:
ok i understand what is going on now, but why do we chose to multiply the numerator and denominator by 2, and why do we chose 2 rather than any other number.

Multiplying by 2 allows the fraction to be rewritten so that part of it is in the form du/u, which can be integrated to to get ln(abs(u))

-GeoMike-
 
trajan22 said:
ok i understand what is going on now, but why do we chose to multiply the numerator and denominator by 2, and why do we chose 2 rather than any other number.

Why don't you tell us what you arrive at after using some other number/method? If it's more effective, we'll ditch the earlier one. :smile:

At an glance, you might think that this is solvable by the substitution u = x2- x +1, but once you do it, you'll realize where the 2 comes from.
 
Last edited:

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 11 ·
Replies
11
Views
1K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 22 ·
Replies
22
Views
3K
Replies
4
Views
2K