Integrating [1/(x^4 - x)]dx

  • Thread starter n4rush0
  • Start date
  • #1
14
0

Homework Statement


[1/(x^4 - x)]dx


Homework Equations





The Attempt at a Solution


I factored the denominator to x(x-1)(x^2 + x +1) and I'm not sure if I can use partial fractions.
 

Answers and Replies

  • #2
Gib Z
Homework Helper
3,346
5
Welcome to PF n4rush0.

What exactly is preventing you from carrying out partial fractions? That is the correct way to proceed.
 
  • #3
14
0
So I set

1/[y(y-1)(y^2+y+1)] = A/y + B/(y-1) + (Cy+D)/(y^2+y+1)

1 = (y-1)(y^2+y+1)A + y(y^2+y+1)B + y(y-1)(Cy+D)

y = 0, A = -1
y = 1, B = 1/3

Not sure what to do for C and D
 
  • #4
dextercioby
Science Advisor
Homework Helper
Insights Author
13,048
595
You're doing something wrong. Just write the whole polynomial in y and equal it to 1. You should get a system of 4 equations with 4 unknowns which is very easy to solve.

EDIT: You're doing it right, but it's a much more tedious work towards the end than in the method I suggested.
 
Last edited:
  • #5
Gib Z
Homework Helper
3,346
5
You're doing something wrong. Just write the whole polynomial in y and equal it to 1. You should get a system of 4 equations with 4 unknowns which is very easy to solve.

No he's doing it right, it's just a different method. However...

n4rush0: Do you notice the pattern of your substitutions? It's to get some numbers in front of just 1 term, and zero's in front of the rest, right? To find C and D, you need to substitute in the zeros of y^2+y+1. You'll end up with a pair of simultaneous equations for C and D, which you can solve.

However, in this case, since you already know A and B, it's easier at that point of the problem to just sub in simple values for y, say -1 and 2, and solve those.

Alternatively, you can do what bigubau suggested.
 
  • #6
34,884
6,624
So I set

1/[y(y-1)(y^2+y+1)] = A/y + B/(y-1) + (Cy+D)/(y^2+y+1)

1 = (y-1)(y^2+y+1)A + y(y^2+y+1)B + y(y-1)(Cy+D)

y = 0, A = -1
y = 1, B = 1/3

Not sure what to do for C and D
Out of curiosity, why did you switch from x that was used in your integral to y here?
 
  • #7
14
0
I tried multiplying it out and got
1 = y^3 (A+B+C) + y^2 (B-C+D) + y(B-D) - A

Even knowing that A = -1 and B = 1/3, I'm still not seeing the "nice" solution.

Oh and, my original problem involved using y, but I wanted to use Mathmatica which solves dx integrals and I copied and pasted [1/(x^4 - x)]dx from there.
 
  • #8
dextercioby
Science Advisor
Homework Helper
Insights Author
13,048
595
Well, then you've missed the most important point in the partial fraction decomposition method.

on the LHS of what you wrote there's supposed to be the following polynomial in y

0 y^3 + 0 y^2 + 0y +1.

Does that help you solve the problem ?
 
  • #9
14
0
Yes, that helps alot. Thanks, I solved it.
 

Related Threads on Integrating [1/(x^4 - x)]dx

  • Last Post
Replies
10
Views
37K
  • Last Post
Replies
5
Views
3K
Replies
7
Views
85K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
5
Views
5K
  • Last Post
Replies
13
Views
185K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
8
Views
23K
  • Last Post
Replies
17
Views
43K
  • Last Post
Replies
5
Views
8K
Top