Question

How do you integrate `int 1/(x^3 -1)` using partial fractions?

Answer 1

Separate into integrable parts.

Explanation:

Let

`I=int1/(x^3-1)dx`

Factorize:

`I=int1/((x-1)(x^2+x+1))dx`

Apply partial fraction decomposition:

`I=1/3int(1/(x-1)-(x+2)/(x^2+x+1))dx`

Rearrange:

`I=1/3int(1/(x-1)-1/2(2x+1)/(x^2+x+1)-3/2 1/(x^2+x+1))dx`

Complete the square in the last term:

`I=1/3int(1/(x-1)-1/2(2x+1)/(x^2+x+1)-6/((2x+1)^2+3))dx`

Integrate term by term:

`I=1/3{ln|x-1|-1/2ln|x^2+x+1|+sqrt3tan^(-1)((2x+1)/sqrt3)}+C`

Simplify:

`I=1/3ln|x-1|-1/6ln|x^2+x+1|+1/sqrt3tan^(-1)((2x+1)/sqrt3)+C`