\[I = \int_1^2 \frac{{\sqrt[3]{{x - {x^3}}}}}{{{x^4}}}dx =\int_1^2 \frac{1}{{{x^3}}}\sqrt[3]{{\frac{{x - {x^3}}}{{{x^3}}}}}dx = \int_1^2 \frac{1}{{{x^3}}}\sqrt[3]{{\frac{1}{{{x^2}}} - 1}}\,dx\]
Đặt \[t = \frac{1}{{{x^2}}} - 1 \Rightarrow dt = \frac{{ - 2}}{{{x^3}}}dx\]
Đổi cận $x=2\Rightarrow t=\dfrac{-3}{4}$
$x=1\Rightarrow t=0$
Vậy \[I = \frac{{ - 1}}{2}\int_0^{\frac{{ - 3}}{4}} \sqrt[3]{t}dt = \frac{{ - 1}}{2}\int_0^{\frac{{ - 3}}{4}} {t^{\frac{1}{3}}}dt = \frac{{ - 3}}{8}\sqrt[3]{{{t^4}}}|_0^{\frac{{ - 3}}{4}} = \frac{{ - 3}}{8}\sqrt[3]{{\frac{{81}}{{256}}}}\]