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

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Answer 1 · 2018-05-22T06:45:31

Separate into integrable parts.

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#