# What rule have they used to change the integral?

In summary, the integral was changed by using the technique of adding zero, where the expression was rewritten in a way that allowed for a cancellation of terms. This is a common trick used in integration, along with other techniques such as u-substitution.

What rule have they used to change the integral from 2x/(x+1)^2?

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.

What rule have they used to change the integral from 2x/(x+1)^2?
I assume you mean 2x/(x-1)^2.
Are you asking how it's valid (isn't it obviously valid?) or how they thought to do that?

eumyang said:
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.

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".

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.

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.

1 person

## 1. What is an integral and why is it important in mathematics?

An integral is a mathematical concept used to calculate the area under a curve or the accumulation of a quantity over a given interval. It is important because it allows us to solve a wide range of problems in various fields such as physics, engineering, and economics.

## 2. How can integrals be changed or manipulated?

Integrals can be changed or manipulated using different rules such as the power rule, product rule, quotient rule, and chain rule. These rules help us simplify the integral and make it easier to solve.

## 3. Why do we need to change integrals?

We change integrals to make them easier to solve or to find an equivalent expression that is more useful for solving a particular problem. Sometimes, changing an integral can also help us to better understand the underlying concepts and relationships.

## 4. What are some commonly used rules for changing integrals?

Some commonly used rules for changing integrals include the power rule, substitution rule, integration by parts, trigonometric substitutions, and partial fraction decomposition. These rules are based on algebraic and trigonometric identities that help us simplify the integrand.

## 5. Are there any restrictions or limitations when changing integrals?

Yes, there can be restrictions or limitations when changing integrals. For example, some integrals may not have an elementary solution and may require advanced techniques such as numerical methods or approximation. Also, when using substitution, we must ensure that the new variable is a valid substitution for the original variable.

Replies
14
Views
1K
Replies
3
Views
1K
Replies
10
Views
1K
Replies
2
Views
791
Replies
3
Views
918
Replies
15
Views
1K
Replies
5
Views
935
Replies
2
Views
1K
Replies
2
Views
1K
Replies
12
Views
1K