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(A) We have $I = \int \frac{dx}{\sqrt{x^2(1-x^3)}} = \int \frac{dx}{x\sqrt{1-x^3}}$.Multiply numerator and denominator by $x^2$:$I = \int \frac{x^2 dx

Vedclass · Accepted Answer

$\frac{(\sqrt{8}-\sqrt{7})}{(\sqrt{8}+\sqrt{7})}$

Vedclass · Answer

(A) We have $I = \int \frac{dx}{\sqrt{x^2(1-x^3)}} = \int \frac{dx}{x\sqrt{1-x^3}}$.Multiply numerator and denominator by $x^2$:$I = \int \frac{x^2 dx}{x^3\sqrt{1-x^3}}$.Let $t = \sqrt{1-x^3}$,then $t^2 = 1-x^3$,so $2t dt = -3x^2 dx$,which means $x^2 dx = -\frac{2}{3}t dt$.Also $x^3 = 1-t^2$.Substituting these into the integral:$I = \int \frac{-\frac{2}{3}t dt}{(1-t^2)t} = -\frac{2}{3} \int \frac{dt}{1-t^2} = -\frac{2}{3} \cdot \frac{1}{2} \log \left| \frac{1+t}{1-t} \right| + C = \frac{1}{3} \log \left| \frac{1-t}{1+t} \right| + C$.Substituting $t = \sqrt{1-x^3}$ back:$I = \frac{1}{3} \log \left| \frac{1-\sqrt{1-x^3}}{1+\sqrt{1-x^3}} \right| + C$.Thus,$f(x) = \frac{1-\sqrt{1-x^3}}{1+\sqrt{1-x^3}}$.For $x = \frac{1}{2}$,$x^3 = \frac{1}{8}$,so $1-x^3 = \frac{7}{8}$.$f\left(\frac{1}{2}\right) = \frac{1-\sqrt{7/8}}{1+\sqrt{7/8}} = \frac{\sqrt{8}-\sqrt{7}}{\sqrt{8}+\sqrt{7}}$.

If $0 < x < 1$ and $\int \frac{dx}{\sqrt{x^2-x^5}} = \frac{1}{3} \log |f(x)| + C$,then $f\left(\frac{1}{2}\right) = $

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