Question

What is the integral of `int 1/(x^3-1) dx`?

Answer 1

`I=1/3ln|x-1|-1/6ln|x^2+x+1|-1/sqrt3tan^-1((2x+1)/sqrt3)+c`

Explanation:

Here,

`I=int1/(x^3-1)dx`

`=int1/((x-1)(x^2+x+1))dx`

Take,

`M=1/((x-1)(x^2+x+1))=A/(x-1)+(Bx+C)/(x^2+x+1)`

`=>1=A(x^2+x+1)+(Bx+C)(x-1)`

`=>1=x^2(A+B)+x(A+B-C)+(A-C)`

Comparing coefficient of `x^2,x` and constant term both sides

`A+B=0to(1) , A+B-C=0to(2),A-C=1to(3)`

Solving ,we get `A=1/3, B=-1/3, C=-2/3`

So,

`I=int(1/3)/(x-1)dx+int(-1/3x-2/3)/(x^2+x+1)dx`

`=1/3ln|x-1|-1/3int(x+2)/(x^2+x+1)dx`

`=1/3ln|x-1|-1/6int(2x+4)/(x^2+x+1)dx`

`=1/3ln|x-1|-1/6int(2x+1+3)/(x^2+x+1)dx`

`=1/3ln|x-1|-1/6int(2x+1)/(x^2+x+1)dx-3/6int1/(x^2+x+1)dx`

`=1/3ln|x-1|-1/6ln|x^2+x+1|-1/2int1/(x^2+x+1/4+3/4)dx`

`=1/3ln|x-1|-1/6ln|x^2+x+1|-1/2intdx/((x+1/2)^2+(sqrt3/2)^2)`

`=1/3ln|x- 1|-1/6ln(x^2+x+1)-1/2xx1/(sqrt3/2)tan^-1((x+1/2)/(sqrt3/2))+c`

`=1/3ln|x-1|-1/6ln|x^2+x+1|-1/sqrt3tan^-1((2x+1)/sqrt3)+c`