- #1
adelin
- 32
- 0
What rule have they used to change the integral from 2x/(x+1)^2?
What rule have they used to change the integral?
Click For Summary
Homework Help Overview
The discussion revolves around the manipulation of integrals, specifically focusing on the technique of "adding zero" to change the form of the integral, such as transforming 2x/(x+1)^2. Participants are exploring the reasoning behind this technique and its validity within the context of integral calculus.
Discussion Character
- Exploratory, Conceptual clarification, Mathematical reasoning
Approaches and Questions Raised
- Participants discuss the method of adding zero to the numerator as a technique to facilitate integration. There are questions about the validity of this approach and how it is conceptualized. Some participants also mention the difference between manipulating expressions and equations, and the potential for using u-substitution as an alternative method.
Discussion Status
The discussion is active, with various participants sharing insights and techniques related to integral manipulation. While there is no explicit consensus, several productive lines of reasoning are being explored, including the creative use of adding zero and the implications of working with expressions versus equations.
Contextual Notes
Participants are considering the implications of different forms of integrals and the techniques available for manipulation, highlighting the constraints of working with expressions as opposed to equations.
What rule have they used to change the integral from 2x/(x+1)^2?
- #2
eumyang
Homework Helper
- 1,347
- 11
It's called "adding zero." 
I see it more as a trick to eventually introduce a factor that can be canceled. For instance, if the integral was this:
\int \frac{3x}{(x-2)^2} dx
I would subtract and add 6:
= \int \frac{3x - 6 + 6}{(x-2)^2} dx
= \int \frac{3x - 6}{(x-2)^2} dx + \int \frac{6}{(x-2)^2} dx
= \int \frac{3(x-2)}{(x-2)^2} dx + \int \frac{6}{(x-2)^2} dx
... etc.
I see it more as a trick to eventually introduce a factor that can be canceled. For instance, if the integral was this:\int \frac{3x}{(x-2)^2} dx
I would subtract and add 6:
= \int \frac{3x - 6 + 6}{(x-2)^2} dx
= \int \frac{3x - 6}{(x-2)^2} dx + \int \frac{6}{(x-2)^2} dx
= \int \frac{3(x-2)}{(x-2)^2} dx + \int \frac{6}{(x-2)^2} dx
... etc.
- #3
- 42,884
- 10,519
I assume you mean 2x/(x-1)^2.adelin said:
Are you asking how it's valid (isn't it obviously valid?) or how they thought to do that?
- #4
- 15,524
- 769
In particular, they added 0 (which doesn't change anything) to the numerator, but with "zero" written as "2-2". There are lots of very creative ways to add zero.eumyang said:It's called "adding zero."
A related technique (not used here) is to multiply by one, where once again you can be very creative in the way in which you write "one".
- #5
Mark44
Mentor
- 38,106
- 10,661
To add to what D H and eumyang have said, there's a big difference between expressions (such as 2x/(x - 1)2 and equations. If you're working with an equation, there are lots of things you can do to get a new equation with the same solution set as the one you started with. For example, you can multiply both sides by a nonzero number, you can add the same amount to both sides, you can take the log of both sides, etc.
With an expression you are much more limited. One thing you can do is add 0 (in some form) or multiply by 1 (also in some form). You can also factor the expression if that seems useful to do, or expand it, if the situation calls for that operation.
With an expression you are much more limited. One thing you can do is add 0 (in some form) or multiply by 1 (also in some form). You can also factor the expression if that seems useful to do, or expand it, if the situation calls for that operation.
- #6
LCKurtz
Science Advisor
Homework Helper
- 9,567
- 775
To add my 2 cents worth, this type of trick is usually done by someone that has done enough u substitutions to see the "need" for it. But you can just as well do the problem using the u-substitution$$
u = x-1,~x = u+1,~du=dx$$ in the first place giving$$
\int \frac {2(u+1)}{u^2}~du = \int\frac 2 u + \frac 2 {u^2}~du$$leading to the same answer.
u = x-1,~x = u+1,~du=dx$$ in the first place giving$$
\int \frac {2(u+1)}{u^2}~du = \int\frac 2 u + \frac 2 {u^2}~du$$leading to the same answer.
Similar threads
- · Replies 6 ·
- · Replies 14 ·
- · Replies 3 ·
- · Replies 10 ·
- · Replies 3 ·
- · Replies 15 ·
- · Replies 2 ·
- · Replies 2 ·