How do you integrate int 1/(x^3 -1)∫1x3−1 using partial fractions?
1 Answer
May 22, 2018
Separate into integrable parts.
Explanation:
Let
I=int1/(x^3-1)dxI=∫1x3−1dx
Factorize:
I=int1/((x-1)(x^2+x+1))dxI=∫1(x−1)(x2+x+1)dx
Apply partial fraction decomposition:
I=1/3int(1/(x-1)-(x+2)/(x^2+x+1))dxI=13∫(1x−1−x+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∫(1x−1−122x+1x2+x+1−321x2+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∫(1x−1−122x+1x2+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)}+CI=13{ln|x−1|−12ln∣∣x2+x+1∣∣+√3tan−1(2x+1√3)}+C
Simplify:
I=1/3ln|x-1|-1/6ln|x^2+x+1|+1/sqrt3tan^(-1)((2x+1)/sqrt3)+CI=13ln|x−1|−16ln∣∣x2+x+1∣∣+1√3tan−1(2x+1√3)+C
