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

  1. Jun 6, 2015 #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{a}{x-1} + \frac{b}{x+1} + \frac{c}{(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


    \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


    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{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
    Thanks for reading.
  2. jcsd
  3. Jun 6, 2015 #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.
  4. Jun 6, 2015 #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.

    Could you explain to me how to apply it?
  5. Jun 6, 2015 #4
    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.
  6. Jun 6, 2015 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    There are two equivalent partial-fraction expansions of the function
    [tex] f(x) = \frac{1}{(x+a)(x+b)^2}, [/tex]
    [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}
    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.
  7. Jun 6, 2015 #6


    Staff: Mentor

    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.
    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.
    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.
  8. Jun 6, 2015 #7
    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.
  9. Jun 6, 2015 #8


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    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You should read what Ray said in post #5.
  10. Jun 6, 2015 #9
    Unfortunately my book seems extremely disorganized ... no details about this. Only this theorem
    exists in the corner of a page.

    Thanks I read the article in purple math, and see some videos in youtube.
  11. Jun 6, 2015 #10


    User Avatar
    Gold Member

    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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