∫1/ (x^3-1)

1. Jun 15, 2012

arishorts

1. The problem statement, all variables and given/known data
∫1/(x^3-1)

2. Relevant equations
A/(x-1) + (Bx+C)/(x^2+x+1)
1=A(x^2+x+1) + (Bx+C)(x-1)

3. The attempt at a solution
I've solved for all the variables in this...
A= 1/3
B= -1/3
C= -2/3

which gets me:
1/3*Ln|x-1| + ∫(-(x/3)-(2/3))/(x^2+x+1)

I can't figure out how to integrate the second half. Maybe i'm not using the right u (u-sub)? There's probably something really simple i'm not seeing.

2. Jun 15, 2012

algebrat

Re: ∫1/(x^3-1)

What if you had 1/(ax^2+b)? Then you would use trig sub right? If you complete the square, then you'll be closer to this form.

3. Jun 15, 2012

Infinitum

Re: ∫1/(x^3-1)

Hi arishorts! Welcome to PF!!

Try substituting $$u = x^2 + x + 1$$

4. Jun 15, 2012

arishorts

Re: ∫1/(x^3-1)

You already know this... but i get du=2x+1 dx, but the numerator is x+2. is there a transition i'm missing?

5. Jun 15, 2012

algebrat

Re: ∫1/(x^3-1)

sorry, i didn't notice the x factor in the numerator, you should integrate that part spearately, you can use the easier u substitution.

6. Jun 15, 2012

arishorts

Re: ∫1/(x^3-1)

I'm trying this right now, don't see why it shouldn't work. Is there not a simpler way?

Edit:

This is what i'm doing right now:

-(1/3)∫ (x+2) / ((3/4)-(x+(1/2))^2)
Then i'd trig sub...

7. Jun 15, 2012

Infinitum

Re: ∫1/(x^3-1)

Write it as

$$\frac{1}{2} (\frac{2x+1}{x^2+x+1} + \frac{3}{x^2+x+1})$$

And then complete the square.

8. Jun 15, 2012

algebrat

Re: ∫1/(x^3-1)

du/2=(x+1/2)dx, then for the leftover (a constant in the numerator), complete the square then use trig sub.

9. Jun 15, 2012

algebrat

Re: ∫1/(x^3-1)

no simpler way, that's why they teach trig sub

EDIT: Infinitum's last post looks good, use simpler u-sub on term on left, trig sub on right.

10. Jun 15, 2012

arishorts

Re: ∫1/(x^3-1)

how can you do that to the numerator, but not the denominator?

11. Jun 15, 2012

Infinitum

Re: ∫1/(x^3-1)

Is this correct?

$$\frac{a+b}{c+d} = \frac{a+b}{c} + \frac{a+b}{d}$$

Put in some values to be sure

12. Jun 15, 2012

arishorts

Re: ∫1/(x^3-1)

You seem to have a very strong algebraic manipulation background, where as i don't. Been trying to work on it lately. I'm trying all methods mentioned here (trying to make sense of it all).

13. Jun 15, 2012

algebrat

Re: ∫1/(x^3-1)

Infinitum actually multiplied by 2(1/2), rather than (1/2)/(1/2). You see, they distributed the 2 to the numerator, and left the 1/2 out front.

14. Jun 15, 2012

algebrat

Re: ∫1/(x^3-1)

do what to the numerator?

15. Jun 15, 2012

arishorts

Re: ∫1/(x^3-1)

not trying to be an ***...

a=1, b=2, c=3, d=4

3/7 ≠ 3/3 + 3/4

16. Jun 15, 2012

arishorts

Re: ∫1/(x^3-1)

it was edited. the 1/2 wasn't there before or i didn't see it. i see it now. Let me try this.

17. Jun 15, 2012

SammyS

Staff Emeritus
Re: ∫1/(x^3-1)

$\displaystyle -\frac{1}{3}\frac{x+2}{x^2+x+1}=-\frac{1}{3}\frac{1}{2}\frac{2x+4}{x^2+x+1}$
$\displaystyle =-\frac{1}{6}\frac{2x+1+3}{x^2+x+1}$

$\displaystyle =-\frac{1}{6}\left(\frac{2x+1}{x^2+x+1}+\frac{3}{x^2+x+1}\right)$​

18. Jun 15, 2012

arishorts

Re: ∫1/(x^3-1)

Got it! Thanks guys... Give you guys +rep if i could, but there doesn't seem to be that system here.

19. Jun 15, 2012

arishorts

Re: ∫1/(x^3-1)

Just got it. Thanks.