$$\eqalign{
& \int {\frac{{dx}}{{{x^3} - 8}}} = \int {\frac{{dx}}{{(x - 2)({x^2} + 2x + 4)}}} \cr
& \frac{1}{{(x - 2)({x^2} + 2x + 4)}} = \frac{A}{{x - 2}} + \frac{B}{{{x^2} + 2x + 4}} \cr
& x = 0 \Rightarrow - \frac{1}{8} = - \frac{A}{2} + \frac{B}{4} \Leftrightarrow 4A - 2B = 1 \cr
& B = 1 \Rightarrow A = \frac{3}{4} \cr
& \int {\frac{{dx}}{{{x^3} - 8}}} = \frac{3}{4}\int {\frac{{dx}}{{x - 2}}} + \int {\frac{{dx}}{{{x^2} + 2x + 4}}} \cr} $$
tới đây tui hết biết rồi.