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

  • #1
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}
##
Thanks for reading.
 

Answers and Replies

  • #2
535
72
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.
 
  • #3
Thanks for replying !

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?
 
  • #4
535
72
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.
 
  • #5
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
Thanks for replying !

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
[tex] f(x) = \frac{1}{(x+a)(x+b)^2}, [/tex]
namely,
[tex] \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}
[/tex]
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.
 
  • #6
34,265
5,905
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
 
  • #7
16
3
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.
 
  • #8
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,812
1,389
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.
 
  • #9
unless your book is extremely disorganized, they should have discussed this technique in the section where this problem appears.
Unfortunately my book seems extremely disorganized ... no details about this. Only this theorem
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.

There are some links on this subject (Wikipedia, purplemaths, even some videos on youtube) with detailed examples
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.
 
  • #10
phion
Gold Member
175
39
If you can't plug and chug values into the partial fraction variables, use row reduction for an [itex]n x m[/itex] matrix.
 

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