<s>[INST] Integral $\int\frac{1}{1+x^3}dx$ 
Calculate$$\int\frac{1}{1+x^3}dx$$

After calculating the partial fractions I got:
$$\frac{1}{3}\int\frac{1}{x+1}dx+\frac{1}{3}\int\frac{2-x}{x^2-x+1}dx=\frac{1}{3}\ln(x+1)+\frac{1}{3}\int\frac{2-x}{x^2-x+1}dx$$
I have no idea on how to proceed. Am I missing a substitution or something?
 [/INST] Hint: You can break up your second calculated indefinite integral into two components as so:
$$\frac{2-x}{x^2-x+1} = -\frac{1}{2}(\frac{2x -1}{x^2-x+1}) + \frac{3/2}{(x-\frac{1}{2})^2 + 3/4}$$
Note that $x^2-x+1 = (x - \frac{1}{2})^2 + \frac{3}{4}$
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