Let $I = \int\limits_{1/3}^{1} \frac{(x-x^3)^{1/3}}{x^4} dx$ Put $x = \sin\theta \Rightarrow dx = \cos\theta d\theta$ Where $x = 1/3, \sin\theta = 1/3$ $\Rightarrow \theta = \sin^{-1} \frac{1}{3}$ When $x = 1$ , $\sin\theta = 1$ $\Rightarrow \theta = \frac{\pi}{2}$ $\therefore I = \int\limits_{\sin^{-1}\frac{1}{3}}^{\frac{\pi}{2}} \frac{(\sin\theta - \sin^3\theta)^{1/3}}{\sin^4\theta} \cos\theta d\theta$ $= \int\limits_{\sin^{-1}\frac{1}{3}}^{\frac{\pi}{2}} \frac{\sin^{1/3}\theta (1-\sin^2\theta)^{1/3}}{\sin^4\theta} \cos\theta d\theta$ $= \int\limits_{\sin^{-1}\frac{1}{3}}^{\frac{\pi}{2}} \frac{(\sin\theta \cos^2\theta)^{1/3}}{\sin^4\theta} \cos\theta d\theta$ $= \int\limits_{\sin^{-1}\frac{1}{3}}^{\frac{\pi}{2}} \frac{ \sin^{1/3}\theta \cos^{5/3}\theta }{ \sin^2\theta \sin^2\theta } d\theta$$= \int\limits_{\sin^{-1}\frac{1}{3}}^{\frac{\pi}{2}} \frac{ \cos^{5/3}\theta }{ \sin^{5/3}\theta } \csc^2\theta d\theta$Put $\cot\theta = t$ $\Rightarrow -\csc^2\theta d\theta = dt$When $\theta = \sin^{-1} \frac{1}{3}$ $\Rightarrow \sin\theta = \frac{1}{3}$ $\Rightarrow \cot\theta = 2\sqrt{2}$ $\Rightarrow t = 2\sqrt{2} = \sqrt{8}$When $\theta = \frac{\pi}{2}, \cot\theta = 0$ $\Rightarrow t = 0$$\therefore I = \int\limits_{\sqrt{8}}^{0} t^{5/3} (-dt)$ $= \int\limits_{0}^{\sqrt{8}} dt = \left[ \frac{t^{8/3}}{8/3} \right]_{0}^{\sqrt{8}}$$= \frac{3}{8} [\sqrt{8}]^{8/3} = \frac{3}{8} (8)^{8/6}$$= \frac{3}{8} (8)^{4/3} = \frac{3}{8} (16) = 6$