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

In summary, this equation can be simplified by taking the derivative of the original equation with respect to x and then dividing by 6.
  • #1
trajan22
134
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
  • #2
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.
 
  • #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.
 
  • #4
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.
 
  • #5
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-
 
  • #6
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:

What is the process of simplifying complex equations?

Simplifying complex equations involves using mathematical operations such as addition, subtraction, multiplication, and division to reduce the equation to its simplest form. This is done by combining like terms, factoring, and using the distributive property.

Why is it important to simplify complex equations?

Simplifying complex equations allows for easier understanding and manipulation of the equation. It also helps to identify patterns and relationships between different parts of the equation, making it easier to solve and apply in real-world scenarios.

What are some common strategies for simplifying complex equations?

Some common strategies for simplifying complex equations include grouping like terms, using the distributive property, factoring, and cancelling out common factors. It is also important to follow the correct order of operations when simplifying equations.

What are the challenges faced when simplifying complex equations?

Simplifying complex equations can be challenging because it requires a strong understanding of algebraic concepts and properties. It also requires careful attention to detail and following the correct order of operations. Additionally, complex equations can involve multiple variables and can be time-consuming to simplify.

How can simplifying complex equations be applied in real life?

Simplifying complex equations can be applied in various fields such as physics, engineering, and economics. In these fields, complex equations are used to model and solve real-world problems. By simplifying these equations, we can better understand and analyze the relationships and patterns within the problem, leading to more accurate and efficient solutions.

Similar threads

  • Calculus and Beyond Homework Help
Replies
11
Views
233
  • Calculus and Beyond Homework Help
Replies
3
Views
268
  • Calculus and Beyond Homework Help
Replies
2
Views
478
  • Calculus and Beyond Homework Help
Replies
6
Views
792
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
634
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
639
  • Calculus and Beyond Homework Help
Replies
4
Views
776
  • Calculus and Beyond Homework Help
Replies
5
Views
190
Back
Top