However, the denominator $x^3+1$ can be factored further using the sum of cubes formula $a^3+b^3 = (a+b)(a^2-ab+b^2)$:
$$x^3+1 = (x+1)(x^2-x+1)$$
So the integral becomes:
$$I = \int \frac{1}{x(x+1)(x^2-x+1)} dx$$
Now, we can set up the partial fraction
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