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

1 Answer
May 22, 2018

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