If int sqrt{x}(1-x^3)^{-frac{1}{2}} dx = frac | vedclass

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(B) Let $I = \int \sqrt{x}(1-x^3)^{-\frac{1}{2}} dx$. Substitute $t = x^{\frac{3}{2}}$,then $dt = \frac{3}{2} x^{\frac{1}{2}} dx$,which implies $\sqrt

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$f(x)=x^{\frac{3}{2}}, g(x)=\sin^{-1} x$

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(B) Let $I = \int \sqrt{x}(1-x^3)^{-\frac{1}{2}} dx$. Substitute $t = x^{\frac{3}{2}}$,then $dt = \frac{3}{2} x^{\frac{1}{2}} dx$,which implies $\sqrt{x} dx = \frac{2}{3} dt$. Since $x^3 = (x^{\frac{3}{2}})^2 = t^2$,the integral becomes: $I = \int (1-t^2)^{-\frac{1}{2}} \left(\frac{2}{3} dt\right) = \frac{2}{3} \int \frac{1}{\sqrt{1-t^2}} dt$. Integrating,we get $I = \frac{2}{3} \sin^{-1}(t) + c$. Substituting $t = x^{\frac{3}{2}}$ back,we have $I = \frac{2}{3} \sin^{-1}(x^{\frac{3}{2}}) + c$. Comparing this with $\frac{2}{3} g(f(x)) + c$,we identify $f(x) = x^{\frac{3}{2}}$ and $g(x) = \sin^{-1} x$.

If $\int \sqrt{x}(1-x^3)^{-\frac{1}{2}} dx = \frac{2}{3} g(f(x)) + c$,then

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