# ∫1/ (x^3-1)

∫1/(x^3-1)

## Homework Equations

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

## 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.

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.

Hi arishorts! Welcome to PF!!

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.

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

Hi arishorts! Welcome to PF!!

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

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

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

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.

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...

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

Write it as

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

And then complete the square.

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

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

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...

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.

Write it as

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

And then complete the square.

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

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

Is this correct?

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

Put in some values to be sure

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

EDIT: it's up to you if you want to consider the x term in the numerator while doing trig sub, I was thinking a constant would be all you'd want to consider. the part with an x in the numerator, i usually go for the much "simpler" u-sub.

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).

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

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.

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

do what to the numerator?

Is this correct?

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

Put in some values to be sure

not trying to be an ***...

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

3/7 ≠ 3/3 + 3/4

do what to the numerator?

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

SammyS
Staff Emeritus
Homework Helper
Gold Member

Write it as

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

And then complete the square.

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

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

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

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

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

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

Just got it. Thanks.