Notice that you may transform the sum of cubes `x^3+1` into a product such that:

`x^3+1 = (x+1)(x^2-x+1)`

You need to use partial fraction decomposition to make the process of evaluation of initial integral easier, such that:

`1/(x^3+1) = A/(x+1) + (Bx+C)/(x^2-x+1)`

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

Opening the brackets yields:

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

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

Equating the coefficients of like powers yields:

`A+B = 0 => A=-B`

`-A+B+C = 0 => -2A=-C`

`A+C = 1 => 3A = 1 => A = 1/3 => C = 2/3 => B = -1/3`

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

Integrating both sides yields:

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

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

You should solve `int x/(x^2-x+1) dx` using substitution `x^2-x+1 = t` such that:

`x^2-x+1 = t => (2x - 1)dx = dt`

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

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

`int x/(x^2-x+1) dx = int (dt)/t + int 1/(x^2-x+1) dx `

`-int x/(x^2-x+1) dx = -int (dt)/t - int 1/(x^2-x+1) dx `

`- int (x)/(x^2-x+1)dx + 2int 1/(x^2-x+1) dx = -int (dt)/t - int 1/(x^2-x+1) dx + 2int 1/(x^2-x+1) dx`

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

You should complete the square `x^2-x` such that:

`x^2-x = x^2-2*(1/2)x + 1/4 - 1/4`

`x^2-x = (x-1/2)^2 - 1/4`

`x^2-x+1 = (x-1/2)^2 - 1/4 + 1`

`x^2-x+1 = (x-1/2)^2 + 3/4`

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

You should solve the integral `int 1/((x-1/2)^2 + 3/4) dx` using the substitution `x-1/2 = v => dx = dv` such that:

`int 1/((x-1/2)^2 + 3/4) dx = 2/sqrt3*arctan(2(x-1/2)/sqrt3)`

`int 1/(x^3+1) dx = (1/3)ln|x+1| - (1/3)ln(x^2-x+1) + 2sqrt3/9*arctan(2(x-1/2)/sqrt3) + c`

**Hence, evaluating the given integral yields `int 1/(x^3+1) dx = (1/3)ln|x+1| - (1/3)ln(x^2-x+1) + 2sqrt3/9*arctan(2(x-1/2)/sqrt3) + c. ` **