Evaluate: ∫ (dx)/(x³ + 1)

$x^3 + 1 = (x + 1)(x^2 - x + 1)$. Write $\dfrac{1}{x^3 + 1} = \dfrac{A}{x + 1} + \dfrac{Bx + C}{x^2 - x + 1}$; then $A = \dfrac{1}{3},\ B = -\dfrac{1}{3},\ C = \dfrac{2}{3}$. $\displaystyle\int \dfrac{dx}{x^3 + 1} = \dfrac{1}{3}\ln|x + 1| - \dfrac{1}{6}\int \dfrac{2x - 1 - 3}{x^2 - x + 1}dx = \dfrac{1}{3}\ln|x + 1| - \dfrac{1}{6}\ln(x^2 - x + 1) + \dfrac{1}{\sqrt{3}}\tan^{-1}\dfrac{2x - 1}{\sqrt{3}} + C$