Evaluate: `intx/(x^3-1)\ dx`

Explanation

`intx/(x^3-1)\ dx`
`=intx/((x-1)(x^2+x+1))dx`
`x/((x-1)(x^2+x+1))=A/(x-1)+(Bx+C)/(x^2+x+1)`
`=>x=A(x^2+x+1)+(Bx+C)(x-1)....(1)`
by solving this we get,
`A=1/3 , C=1/3` and `B=-1/3`
now integration becomes,
`I=intA/(x-1)+(Bx+C)/(x^2+x+1)dx`
`=1/3 int 1/(x-1)+int (-1/3x+1/3)/(x^2+x+1)dx`
`=1/3 ln (x-1)-1/3int (x-1)/(x^2+x+1)dx`
taking `I_1=int (x-1)/(x^2+x+1)dx`
put `x^2+x+1=t`
`=>dt=(2x+1)dx`
putting these values we get,
`I_1=1/2int (2(x-1)+3-3)/(x^2+x+1)dx`
`=1/2 int (dt)/t-1/2*3 int1/((x+1/2)^2+3/4)dx`
`=1/2 ln(x^2+x+1)-3/2( tan^-1 (x+1/2))/(sqrt3/2
)`
`=1/2 ln (x^2+x+1)-sqrt3tan^-1(2x+1)/sqrt3+c`
now,`I=1/3 ln(x+1)-1/3 (1/2 ln|x^2+x+1|-sqrt3 tan^-1 (2x+1)/sqrt3)+c`
`=1/3 ln|x+1|-1/6 ln|x^2+x+1|+1/sqrt3tan^-1(2x+1)/sqrt3+c`