1. ## Partial fractions (integration)

Evaluate the integral:

$\int \frac{1}{x^3 - 1} dx$

I got to the final step:

$=ln|3x - 1| - \int \frac{3x}{x^2 + x + 1}dx + \int \frac{\frac{2}{3}}{x^2 + x + 1} dx$

but dont know how to integrate the 2 integrals due to the irreducible quadratic factors of the denominator.

2. What did you get for your partial-fraction decomposition? Because I got
$\dfrac{1/3}{x - 1} - \dfrac{(1/3)x + 2/3}{x^2 + x + 1}$
$\dfrac{1}{3(x - 1)} - \dfrac{x + 2}{3(x^2 + x + 1)}$
but it doesn't look like it would lead to your final step.

3. The procedure of integrate the above is quite standard. I don't like to use wolfram in many occasions, but this is an exception...
int 1&#47;&#40;X&#94;3-1&#41; - Wolfram|Alpha

(Press on 'show steps')

4. Originally Posted by SyNtHeSiS
Evaluate the integral:

$\int \frac{1}{x^3 - 1} dx$

I got to the final step:

$=ln|3x - 1| - \int \frac{3x}{x^2 + x + 1}dx + \int \frac{\frac{2}{3}}{x^2 + x + 1} dx$

but dont know how to integrate the 2 integrals due to the irreducible quadratic factors of the denominator.
When I split your original integral into partial fractions, I get

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

since

$(x-1)\left(x^2+bx+c\right)=x^3-1$

$x^3+bx^2+cx-x^2-bx-c=x^3-1$

$b=1,\ c=1$

Then

$\displaystyle\huge\frac{1}{(x-1)\left(x^2+x+1\right)}=\frac{A}{x-1}+\frac{B+Cx}{x^2+x+1}$

giving

$A=\frac{1}{3},\ B=-\frac{2}{3},\ C=-\frac{1}{3}$

5. Originally Posted by SyNtHeSiS
Evaluate the integral:

$\int \frac{1}{x^3 - 1} dx$

I got to the final step:

$=ln|3x - 1| - \int \frac{3x}{x^2 + x + 1}dx + \int \frac{\frac{2}{3}}{x^2 + x + 1} dx$

but dont know how to integrate the 2 integrals due to the irreducible quadratic factors of the denominator.
We should note that

$\int \frac{1}{x^3 - 1} dx = \int \frac{1}{x^3 - 1^3} dx$

And,

$x^3 - 1^3 = (x-1)(x^2 +x + 1)$

Resulting in,

$\int \frac{1}{ (x-1)(x^2 +x + 1) } dx$

Partial Fractions yields

$\frac{ A }{x-1} + \frac{ Bx + C }{ (x^2 +x + 1) }$

$\frac{ A (x^2 +x + 1) + B(x^2 - x) + C(x-1) } { (x-1)(x^2 +x + 1) }$

$1: A - C = 1$

$x: A -B + C = 0$

$x^2: A + B = 0$

This yields,

$C = - \frac{2}{3}$

$A = \frac{1}{3}$

$B = - \frac{1}{3}$

$\int \frac{1}{ (x-1)(x^2 +x + 1) } dx = \frac{1}{3} \int \frac{1}{x-1}dx -\frac{1}{3} \int \frac{x + 2}{(x^2 +x + 1) }dx$

The second integral can be evaluated as follows,

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

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

Now let's look at

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

What happens when we try to make this look like the inverse tan integral?

$\frac{2}{3} \int \frac{4}{(2x+1)^2 + ( \sqrt{3} )^2 }dx$

If you expand inside the integral (to check that we have done this correctly) we will see that

$\frac{4}{(2x+1)^2 + ( \sqrt{3} )^2 } = \frac{4}{ 4x^2 + 4x + 4 } = \frac{1}{(x^2 +x + 1) }$

It should be obvious then that,

$\frac{2}{3} \int \frac{1}{(x^2 +x + 1) }dx = \frac{8}{3} \int \frac{1}{(2x+1)^2 + ( \sqrt{3} )^2 }dx = \frac{8}{3 \sqrt{3} } tan^{-1} ( \frac{ 2x+1}{ \sqrt{3} }) + C$

I leave you to consider

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