How do you find the integral of 1/(x^3-1) dx?

Question

Dont need the answer just steps to make it easy

Answer 1

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

Split the fraction into partial fractions. You will get one term A/(x-1) which you can integrate to get Aln|x-1|.

The other term (Bx + C)/(x^2 + x + 1) is a bit trickier.

(2x + 1)/(x^2 + x + 1) can be integrated to get ln|x^2 + x + 1|. You will be left with a term D/(x^2 + x + 1)

Remember that 1/(x^2 + a^2) can be integrated by (1/a)arctan(x/a).

D/(x^2 + x + 1)
= D/(x^2 + x + 1/4 + 3/4)
= D/[(x+1/2)^2 + ((1/2)sqrt(3))^2)

Let u = x + 1/2
du = dx

You should be able to integrate this from here onwards.


You can always check your answer on the attached link.

Answer 2

When the denominator can be factored, please try to do it:
[integral sign] 1/(x^3-1) dx
= [integral sign] (1/3){1/(x -1) - (x + 2)/(x^2+x+1)} dx
= [integral sign] {(1/3)/(x -1) - (1/6)(2x+1 +3)/(x^2+x+1)} dx
I think that you may know how to do it from this point on.
Good Luck!