$\int\limits \frac{1}{x^3+1}dx=\frac{1}{3}\int\limits( \frac{1}{x+1}-\frac{x-2}{x^2-x+1})dx$ $=\frac{1}{3}\int\limits[ \frac{1}{x+1}-\frac{2x-1}{2(x^2-x+1)}+\frac{3}{2(x^2-x+1)}]dx$ $=\frac{1}{3}\int\limits[ \frac{1}{x+1}-\frac{2x-1}{2(x^2-x+1)}+\frac{3}{2}.\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}]dx$ $\frac{1}{3}[ln|x+1|-\frac{1}{2}ln(x^2-x+1)+\sqrt{3}arctan(\frac{2x-1}{\sqrt{3}})]+C$

$\int\limits \frac{1}{x^3+1}dx=\frac{1}{3}\int\limits( \frac{1}{x+1}-\frac{x-2}{x^2-x+1})dx$ $=\frac{1}{3}\int\limits[ \frac{1}{x+1}-\frac{2x-1}{2(x^2-x+1)}+\frac{3}{2(x^2-x+1)}]dx$ $=\frac{1}{3}\int\limits[ \frac{1}{x+1}-\frac{2x-1}{2(x^2-x+1)}+\frac{3}{2}.\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}]dx$ $=\frac{1}{3}[ln|x+1|-\frac{1}{2}ln(x^2-x+1)+\sqrt{3}arctan(\frac{2x-1}{\sqrt{3}})]+C$