- #1

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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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- Thread starter n4rush0
- Start date

- #1

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

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

- #2

Gib Z

Homework Helper

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What exactly is preventing you from carrying out partial fractions? That is the correct way to proceed.

- #3

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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

- 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.

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

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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

Mark44

Mentor

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Out of curiosity, why did you switch from x that was used in your integral to y here?

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

- #7

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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

- 13,048

- 595

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

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Yes, that helps alot. Thanks, I solved it.

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