Integrating a Rational Function with a Quadratic Denominator

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Homework Help Overview

The discussion revolves around integrating a rational function with a quadratic denominator, specifically the integral of 1/(x^4 - x)dx. Participants are exploring the method of partial fractions for this integration.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to factor the denominator and apply partial fractions but expresses uncertainty about the process. Some participants question what is preventing the application of partial fractions and suggest methods for determining coefficients in the decomposition. Others discuss the implications of variable substitution and the structure of the polynomial formed during the decomposition.

Discussion Status

Participants are actively engaging with the problem, offering guidance on how to set up equations for the coefficients in the partial fraction decomposition. There is a recognition of different methods being discussed, and while some participants express confusion, others provide clarifications that seem to aid understanding.

Contextual Notes

There is mention of using different variables (x and y) in the context of the integral and the partial fraction decomposition, which raises questions about consistency in notation. The original problem context and the use of software tools like Mathematica are also noted.

n4rush0
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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.
 
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Welcome to PF n4rush0.

What exactly is preventing you from carrying out partial fractions? That is the correct way to proceed.
 
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
 
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:
bigubau said:
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.
 
n4rush0 said:
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?
 
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.
 
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 ?
 
Yes, that helps a lot. Thanks, I solved it.
 

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