1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Partial Fractions

  1. Dec 30, 2007 #1

    There's this one problem that I'm having troubles with. I've tried using the decomposition method, but I've ended up getting a messy answer. If someone can give me tips or the solution to the problem, I'll appreciate it. Here's the problem: solve the integral of 1/ y^2-1 dx.
  2. jcsd
  3. Dec 30, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    I assume you mean [tex]\int\frac{1}{x^2-1}dx[/tex].

    So if you want to use partial fractions, you can decompose it as such:
    [tex]\frac{1}{x^2-1}=\frac{A}{x+1}+\frac{B}{x-1}[/tex]. How would you continue?
  4. Dec 30, 2007 #3
    I used that method however it came out pretty ugly.

    (1) A/ y+1 + B/ y-1

    (2) 1 = (y+1) (y-1) A/ y+1 + (y+1) (y-1) B/ y-1 -I multiplied each side by the LCD

    (3) 1 = A(y-1) +B(y+1) -Cancelled algebraically

    (4) 1 = y(A+B) + (-A+B)

    Since there's only one constant, am I assume that (A+B) =1 and similarly that -A+B =1? In order to identify A and B, I set up the system of equations as:

    A+B =1
    -A+B= 1

    I know there's something wrong here because for any subsitution I would get "0" Plus, I've never seen a partial decomp problem where A and B would equal to the same number.
    Last edited: Dec 30, 2007
  5. Dec 30, 2007 #4


    User Avatar
    Science Advisor
    Homework Helper

    A little mistake at the end. Since there is no y term on the left side, you require A+B = 0, while -A + B = 1.
  6. Dec 30, 2007 #5
    Thanks Nicksauce,

    The only rational answer I can see why A+B can equal to 0 is because the costants on both sides of the equation are equal somehow

    Here's what I got from using that helpful tip

    (1) 1/ y^2-1 = A/ y-1 + B/ y+1

    (2) -1/2/ y-1 + 1/2/ y+1

    (3) integral 1/ y^2-1 dx = integral - dx/ 2(y-1) + dx/ 2(y+1)

    (4) -1/2 integral dx/ y-1 + 1/2 integral dx/ y+1

    = 1/2 ln (y+1) -1/2 ln (y-1) whew!

    = 1/2 ln ((y-1)/(y+1))

    Thanks for the help! :-)
    Last edited: Dec 30, 2007
  7. Dec 30, 2007 #6


    User Avatar
    Science Advisor
    Homework Helper

    No problemo.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Partial Fractions
  1. Partial Fractions (Replies: 2)

  2. Partial fractions (Replies: 2)

  3. Partial fraction (Replies: 4)

  4. Partial fraction (Replies: 3)