The value of the integral int_{frac{1}{3}}^{1 | vedclass

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(A) Let $I = \int_{\frac{1}{3}}^{1} \frac{(x-x^3)^{\frac{1}{3}}}{x^4} dx$.We can rewrite the integrand as:$I = \int_{\frac{1}{3}}^{1} \frac{x(1-x^2)^{

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$6$

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(A) Let $I = \int_{\frac{1}{3}}^{1} \frac{(x-x^3)^{\frac{1}{3}}}{x^4} dx$.We can rewrite the integrand as:$I = \int_{\frac{1}{3}}^{1} \frac{x(1-x^2)^{\frac{1}{3}}}{x^4 \cdot x^{\frac{1}{3}}} dx = \int_{\frac{1}{3}}^{1} \frac{(1-x^2)^{\frac{1}{3}}}{x^3 \cdot x^{\frac{1}{3}}} dx = \int_{\frac{1}{3}}^{1} \frac{(1-x^2)^{\frac{1}{3}}}{x^{\frac{10}{3}}} dx$.Let $1-x^2 = t x^2$. Then $1 = (t+1)x^2$,so $x^2 = \frac{1}{t+1}$.Differentiating both sides: $2x dx = -\frac{1}{(t+1)^2} dt$,so $dx = -\frac{1}{2x(t+1)^2} dt$.Also,$x = (t+1)^{-\frac{1}{2}}$,so $dx = -\frac{1}{2(t+1)^{-\frac{1}{2}}(t+1)^2} dt = -\frac{1}{2(t+1)^{\frac{3}{2}}} dt$.When $x = \frac{1}{3}$,$t = \frac{1-x^2}{x^2} = \frac{1 - 1/9}{1/9} = 8$.When $x = 1$,$t = 0$.Substituting into the integral:$I = \int_{8}^{0} \frac{(t x^2)^{\frac{1}{3}}}{x^4} \left( -\frac{1}{2x(t+1)^2} \right) dt = \int_{0}^{8} \frac{t^{\frac{1}{3}} x^{\frac{2}{3}}}{x^5} \cdot \frac{1}{2(t+1)^2} dt$.Since $x^2 = (t+1)^{-1}$,$x = (t+1)^{-\frac{1}{2}}$,so $x^4 = (t+1)^{-2}$.$I = \int_{0}^{8} \frac{t^{\frac{1}{3}} (t+1)^{-\frac{1}{3}}}{(t+1)^{-2}} \cdot \frac{1}{2(t+1)^2} dt = \frac{1}{2} \int_{0}^{8} t^{\frac{1}{3}} (t+1)^{-\frac{1}{3} + 2 - 2} dt = \frac{1}{2} \int_{0}^{8} t^{\frac{1}{3}} (t+1)^{-\frac{1}{3}} dt$.This approach is complex. Let's use $x = \frac{1}{u}$. Then $dx = -\frac{1}{u^2} du$.$I = \int_{3}^{1} \frac{(\frac{1}{u} - \frac{1}{u^3})^{\frac{1}{3}}}{1/u^4} (-\frac{1}{u^2}) du = \int_{1}^{3} u^4 \frac{(\frac{u^2-1}{u^3})^{\frac{1}{3}}}{u^2} du = \int_{1}^{3} u^2 \frac{(u^2-1)^{\frac{1}{3}}}{u} du = \int_{1}^{3} u (u^2-1)^{\frac{1}{3}} du$.Let $u^2-1 = v$,then $2u du = dv$.$I = \frac{1}{2} \int_{0}^{8} v^{\frac{1}{3}} dv = \frac{1}{2} [\frac{3}{4} v^{\frac{4}{3}}]_{0}^{8} = \frac{3}{8} (8^{\frac{4}{3}}) = \frac{3}{8} (16) = 6$.Thus,the correct option is $A$.

The value of the integral $\int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}} d x$ is

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