The value of the integral $\int\limits_{1/3}^1 {\cfrac{{{{\left( {x - {x^3}} \right)}^{1/3}}}}{{{x^4}}}} dx$ is

Option a is correct

Let $I = \int\limits_{\cfrac{1}{3}}^1 {\cfrac{{{{\left( {x - {x^3}} \right)}^{\cfrac{1}{3}}}}}{{{x^4}}}} dx$

When $x = \cfrac{1}{3},\sin \theta = \cfrac{1}{3}$ $\Rightarrow$ $\theta = {\sin ^{ - 1}}\cfrac{1}{3}$ and
When $x = 1,\sin \theta = 1$ $\Rightarrow$ $\theta = \cfrac{\pi }{2}$

$\therefore$ $I = \int\limits_{{{\sin }^{ - 1}}\cfrac{1}{3}}^{\cfrac{\pi }{2}} {\cfrac{{{{\left( {\sin \theta - {{\sin }^3}\theta } \right)}^{\cfrac{1}{3}}}}}{{{{\sin }^4}\theta }}\cos \theta d\theta }$

$= \int\limits_{{{\sin }^{ - 1}}\cfrac{1}{3}}^{\cfrac{\pi }{2}} {\cfrac{{{{\sin }^{\cfrac{1}{3}}}\theta {{\left( {1 - {{\sin }^2}\theta } \right)}^{\cfrac{1}{3}}}}}{{{{\sin }^4}\theta }}\cos \theta d\theta }$

$= \int\limits_{{{\sin }^{ - 1}}\cfrac{1}{3}}^{\cfrac{\pi }{2}} {\cfrac{{{{\left( {\sin \theta {{\cos }^2}\theta } \right)}^{\cfrac{1}{3}}}}}{{{{\sin }^4}\theta }}\cos \theta d\theta }$

$= \int\limits_{{{\sin }^{ - 1}}\cfrac{1}{3}}^{\cfrac{\pi }{2}} {\cfrac{{{{\sin }^{\cfrac{1}{3}}}\theta {{\cos }^{\cfrac{5}{2}}}\theta }}{{{{\sin }^2}\theta {{\sin }^2}\theta }}d\theta }$

$= \int\limits_{{{\sin }^{ - 1}}\cfrac{1}{3}}^{\cfrac{\pi }{2}} {\cfrac{{{{\cos }^{\cfrac{5}{3}}}\theta }}{{{{\sin }^{\cfrac{5}{3}}}\theta }}\cos e{c^2}\theta d\theta }$

When $\theta = {\sin ^{ - 1}}\cfrac{1}{3}$ $\Rightarrow$

$\sin \theta = \cfrac{1}{3}$ $\Rightarrow$ $\cot \theta = 2\sqrt 2$

$\Rightarrow$ $t = 2\sqrt 2 = \sqrt 8$ and

When $\theta = \cfrac{\pi }{2},\cot \theta = 0$ $\Rightarrow$ $t = 0$
$\therefore$ $I = \int\limits_{\sqrt 8 }^0 {{t^{\cfrac{5}{3}}}\left( { - dt} \right) - = \int\limits_0^{\sqrt 8 } {{t^{\cfrac{5}{3}}}dt} }$

$= \left[ {\cfrac{{{t^{\cfrac{8}{3}}}}}{{\cfrac{8}{3}}}} \right]_0^{\sqrt 8 } = \cfrac{3}{8}{\left[ {\sqrt 8 } \right]^{\cfrac{8}{3}}} = \cfrac{3}{8}{\left( 8 \right)^{\cfrac{8}{6}}} = \cfrac{3}{8}{\left( 8 \right)^{\cfrac{4}{3}}} = \cfrac{3}{8}\left( {16} \right) = 6$