How do you integrate int 1/(x^3 -1)1x31 using partial fractions?

1 Answer
May 22, 2018

Separate into integrable parts.

Explanation:

Let

I=int1/(x^3-1)dxI=1x31dx

Factorize:

I=int1/((x-1)(x^2+x+1))dxI=1(x1)(x2+x+1)dx

Apply partial fraction decomposition:

I=1/3int(1/(x-1)-(x+2)/(x^2+x+1))dxI=13(1x1x+2x2+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))dxI=13(1x1122x+1x2+x+1321x2+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))dxI=13(1x1122x+1x2+x+16(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)}+CI=13{ln|x1|12lnx2+x+1+3tan1(2x+13)}+C

Simplify:

I=1/3ln|x-1|-1/6ln|x^2+x+1|+1/sqrt3tan^(-1)((2x+1)/sqrt3)+CI=13ln|x1|16lnx2+x+1+13tan1(2x+13)+C