Evaluate `int 1/(x^3-1)dx` :
`int 1/(x^3-1)dx=int (dx)/((x-1)(x^2+x+1)) `
Use partial fraction decomposition to get:
`=1/3 int (1/(x-1)-(x+2)/(x^2+x+1))dx `
` `` (x+2)/(x^2+x+1)=(1/2)(2x+1+3)/(x^2+x+1) `
` =(1/2)( (2x+1)/(x^2+x+1)+3/((x+1)^2+3/4))`
` "So"`
` =1/3int(dx)/(x-1)-1/6int(x+2)/(x^2+x+1)dx-1/6int3/((x+1)^2+3/4)dx`
`=1/3ln(x-1)-1/6ln(x^2+x+1)-sqrt(3)/3 tan^(-1)(2x+1)/sqrt(3)+C`
`=1/6(2ln(x-1)-ln(x^2+x+1)-2sqrt(3) tan^(-1)((2x+1)/sqrt(3)))+C`
``
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