# ##\int\frac{2x+6}{(x-1)(x+1)^2} dx## ?

• Byeonggon Lee
In summary, the conversation discusses a problem involving integral by substitution and partial fraction decomposition. The book's explanation uses a technique that the speaker finds unintuitive, and they struggle to understand the process. The conversation ends with the speaker asking for clarification on why a certain approach does not work and how they can improve their understanding of this technique.
Byeonggon Lee
Hi I'm currently doing 'integral by substitution' part in a book.

Although it is integral by substitution part, some exercises are solved using reduction of fraction and integral, without substitution.

(Actually I can't solve some exercises if I use substitution and the book's explanation also doesn't use substitution)

By the way I'm stuck with this exercise:

##\int \frac{2x+6}{(x-1)(x+1)^2} dx##

I tried to use substitution but that makes the problem more complicated and I can't also use reduction of fraction.

So I looked up the book's explanation, the book solved this problem in a weird way :

##
\frac{2x+6}{(x-1)(x+1)^2}
= \frac{a}{x-1} + \frac{b}{x+1} + \frac{c}{(x+1)^2}
=\frac{a(x+1)^2+b(x-1)(x+1)+c(x-1)}{(x-1)(x+1)^2}
##

##
2x+6 = (a+b)x^2+(2a+c)x+a-b-c
##

##
a+b=0, 2a+c=2, a-b-c=6
##

##
a=2,b=-2,c=-2
##

##
\int \frac{2x+6}{(x-1)(x+1)^2} dx = \int (\frac{2}{x-1}-\frac{2}{x+1}-\frac{2}{(x+1)^2})dx
##

##
=2\ln|x-1|-2\ln|x+1|+\frac{2}{x+1}+C
##

The book didn't introduced technique like this.
And I feel unnatural and not intuitive about its process because..
Who would imagine to split the expression

##
\frac{2x+6}{(x-1)(x+1)^2}
##

into

##
\frac{a}{x-1} + \frac{b}{x+1} + \frac{c}{(x+1)^2}
##

I can't even use only a, b instead of a, b, c:

##
\frac{2x+6}{(x-1)(x+1)^2} = \frac{a}{x-1}+\frac{b}{x+1}=\frac{a(x+1)^2+b(x^2-1)}{(x-1)(x+1)^2}=\frac{(a+b)x^2+2ax+a-b}{(x-1)(x+1)^2}
##

##
a+b=0, 2a=2,a-b=6
##

And that is contradiction 'a' should be 1 because 2a=2
and b should be -5 because a-b=6
but that doesn't make a+b=0

I don't know why this doesn't work.

So my question is

1. why last thing (use only a, b instead of a, b, c) doesn't work?

2. How can I think this way immediately when I see an exercise like this? Is this technique frequently used?? and how can I think to split the expression into exactly a,b,c, not a, b
split the expression
##
\frac{2x+6}{(x-1)(x+1)^2}
##
into
##
\frac{a}{x-1} + \frac{b}{x+1} + \frac{c}{(x+1)^2}
##

The technique you mention is called partial fraction decomposition.
It is very useful when you try to find primitives of functions that have the form of a fraction of polynomials.
It greatly simplifies the search, breaking a complicated expression, for which a primitive is not easily accessible, into a sum of expressions for which it is usually much simpler to find a primitive.
You should get familiar with this subject that requires some training.

My book also introduced partial fraction decomposition which is
##\frac{1}{AB} =\frac{1}{B-A}(\frac{1}{A}-\frac{1}{B})##

But this following expression is so complicated that I can't apply partial fraction decomposition.
##\frac{2x+6}{(x−1)(x+1)^2}##

Could you explain to me how to apply it?

No this is a particular case of partial fraction decomposition. And it could be an incomplete decomposition depending of A and B.
I cannot give you a decent explanation in a few lines, you should study it though it will take you out of your current focus (it can be time consuming at first).
There are some links on this subject (Wikipedia, purplemaths, even some videos on youtube) with detailed examples.
Maybe you should finish your homework, then study fraction decomposition, and go back to this problem.

Byeonggon Lee said:

My book also introduced partial fraction decomposition which is
##\frac{1}{AB} =\frac{1}{B-A}(\frac{1}{A}-\frac{1}{B})##

But this following expression is so complicated that I can't apply partial fraction decomposition.
##\frac{2x+6}{(x−1)(x+1)^2}##

Could you explain to me how to apply it?

There are two equivalent partial-fraction expansions of the function
$$f(x) = \frac{1}{(x+a)(x+b)^2},$$
namely,
$$\begin{array}{rl} 1: & f(x) = \frac{A_1}{x+a} + \frac{B_1 x + C_1}{(x+b)^2} \\ 2: & f(x) = \frac{A_2}{(x+a)} + \frac{B_2}{(x+b)} + \frac{C_2}{(x+b)^2} \end{array}$$
Here, ##A_1, B_1, C_1## and/or ##A_2, B_2, C_2## are constants. If you Google "partial fractions" you will see several articles that show you exactly how the ##A,B,C## can be obtained.

You might think about why forms 1 and 2 above are equivalent.

Do not worry about the fact that you might not have been able to "discover" such things on your own. All that matters is that somebody else discovered them long ago and that they are well-known and well-documented.

Byeonggon Lee said:
By the way I'm stuck with this exercise:

##\int \frac{2x+6}{(x-1)(x+1)^2} dx##

I tried to use substitution but that makes the problem more complicated and I can't also use reduction of fraction.
Why not? The most natural approach to this problem would be partial fraction decomposition. I don't think there is a substitution that would work.
Byeonggon Lee said:
So I looked up the book's explanation, the book solved this problem in a weird way :

##
\frac{2x+6}{(x-1)(x+1)^2}
= \frac{a}{x-1} + \frac{b}{x+1} + \frac{c}{(x+1)^2}
=\frac{a(x+1)^2+b(x-1)(x+1)+c(x-1)}{(x-1)(x+1)^2}
##

##
2x+6 = (a+b)x^2+(2a+c)x+a-b-c
##

##
a+b=0, 2a+c=2, a-b-c=6
##

##
a=2,b=-2,c=-2
##

##
\int \frac{2x+6}{(x-1)(x+1)^2} dx = \int (\frac{2}{x-1}-\frac{2}{x+1}-\frac{2}{(x+1)^2})dx
##

##
=2\ln|x-1|-2\ln|x+1|+\frac{2}{x+1}+C
##

The book didn't introduced technique like this.
I have a hard time believing this. They used partial fraction decomposition, so unless your book is extremely disorganized, they should have discussed this technique in the section where this problem appears.
Byeonggon Lee said:
And I feel unnatural and not intuitive about its process because..
Who would imagine to split the expression

##
\frac{2x+6}{(x-1)(x+1)^2}
##

into

##
\frac{a}{x-1} + \frac{b}{x+1} + \frac{c}{(x+1)^2}
##

I can't even use only a, b instead of a, b, c:
This is because you have a repeated linear factor, ##(x + 1)^2##. Since you don't have it in your work below, you get a set of numbers that is incorrect.
Byeonggon Lee said:
##
\frac{2x+6}{(x-1)(x+1)^2} = \frac{a}{x-1}+\frac{b}{x+1}=\frac{a(x+1)^2+b(x^2-1)}{(x-1)(x+1)^2}=\frac{(a+b)x^2+2ax+a-b}{(x-1)(x+1)^2}
##

##
a+b=0, 2a=2,a-b=6
##

And that is contradiction 'a' should be 1 because 2a=2
and b should be -5 because a-b=6
but that doesn't make a+b=0

I don't know why this doesn't work.

So my question is

1. why last thing (use only a, b instead of a, b, c) doesn't work?

2. How can I think this way immediately when I see an exercise like this? Is this technique frequently used?? and how can I think to split the expression into exactly a,b,c, not a, b

The last numerator should be in the form c+dx, although d=0 in this case.
Partial fraction is indeed a technique very frequently used and I believe any elementary calculus textbook will introduce it.

momoko said:
The last numerator should be in the form c+dx, although d=0 in this case.
You should read what Ray said in post #5.

Mark44 said:
unless your book is extremely disorganized, they should have discussed this technique in the section where this problem appears.
Byeonggon Lee said:
1AB=1B−A(1A−1B)\frac{1}{AB} =\frac{1}{B-A}(\frac{1}{A}-\frac{1}{B})
exists in the corner of a page.

geoffrey159 said:
There are some links on this subject (Wikipedia, purplemaths, even some videos on youtube) with detailed examples
Ray Vickson said:
1:2:f(x)=A1x+a+B1x+C1(x+b)2f(x)=A2(x+a)+B2(x+b)+C2(x+b)2

Thanks I read the article in purple math, and see some videos in youtube.

If you can't plug and chug values into the partial fraction variables, use row reduction for an $n x m$ matrix.

## 1. What is the purpose of the integral?

The purpose of the integral is to find the area under the curve represented by the given function.

## 2. How do you solve the integral?

To solve the integral, you can use the method of partial fractions to break down the given function into simpler fractions. Then, you can integrate each fraction separately.

## 3. What are the limits of integration?

The limits of integration depend on the specific problem or application. They are typically given in the problem or can be determined based on the context.

## 4. Can the integral be solved using other methods?

Yes, the integral can also be solved using techniques such as substitution, integration by parts, or trigonometric substitution. However, the method of partial fractions is often the most efficient for solving integrals of this form.

## 5. What is the significance of the constants in the partial fraction decomposition?

The constants in the partial fraction decomposition represent the coefficients of the simplified fractions that were used to break down the original function. These constants are important for integrating each fraction separately and finding the final solution.

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