Integrate $ \int \frac{1}{1 + x^3}dx $ $$ \int \frac{1}{1 + x^3}dx  $$
Attempt:
I added and subtracted $x^3$ in the numerator but after a little solving I can't get through.

Does this help?
$\dfrac{1}{x^3+1} = \dfrac{1}{(x+1)(x^2-x+1)} = \dfrac13\left(\dfrac{1}{x+1} - \dfrac{x-2}{x^2-x+1}\right)$
Can you integrate it from there?
The integral of the second part is a little tricky. Note that:
$\dfrac13\left(\dfrac{x-2}{x^2-x+1}\right) = \dfrac16\left(\dfrac{2x-4}{x^2-x+1}\right)= \dfrac16\left(\dfrac{2x-1}{x^2-x+1} - \dfrac{3}{x^2-x+1}\right)$
$=\dfrac16\left(\dfrac{2x-1}{x^2-x+1} - \dfrac{3}{\left(x-\frac12\right)^2 + \frac34}\right)$
That first integral is a $u$-substitution, and the second uses the arctangent function.
