What rule have they used to change the integral?

[adelin]

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

 

[eumyang]

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.

 

[haruspex]

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?

 

[D H]

It's called "adding zero."

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".

 

[Mark44]

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.

 

[LCKurtz]

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.