<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] To integrate $\frac{2- x}{x^2- x+ 1}$, complete the square in the denominator:  $\int \frac{2- x}{x^2- x+ 1}dx= \int\frac{2- x}{x^2- x+ \frac{1}{4}-\frac{1}{4}+ 1}dx= \int\frac{2-x}{\left(x- \frac{1}{2}\right)+ \frac{3}{4}}dx$.
Now let $u= x- \frac{1}{2}$ so that $du= dx$ and $x= u+ \frac{1}{2}$ and $2- x= \frac{3}{2}- u$ and $dx= -dy.
$\int \frac{2- x}{x^2- x+ 1}dx= $$-\int \frac{\frac{3}{2}- u}{u^2+ \frac{3}{4}}du= $$-\frac{3}{2}\int\frac{du}{u^2+ \frac{3}{4}}+ \int\frac{u du}{u^2+ \frac{3}{4}}$.
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