Let int frac{x^{1/2}}{sqrt{1-x^3}} dx = frac{ | vedclass

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(B) We are given the integral $I = \int \frac{x^{1/2}}{\sqrt{1-x^3}} dx$. Let $t = x^{3/2}$. Then $dt = \frac{3}{2} x^{1/2} dx$, which implies $x^{1/2

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

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(B) We are given the integral $I = \int \frac{x^{1/2}}{\sqrt{1-x^3}} dx$. Let $t = x^{3/2}$. Then $dt = \frac{3}{2} x^{1/2} dx$, which implies $x^{1/2} dx = \frac{2}{3} dt$. Substituting these into the integral, we get: $I = \int \frac{\frac{2}{3} dt}{\sqrt{1-t^2}} = \frac{2}{3} \int \frac{dt}{\sqrt{1-t^2}} = \frac{2}{3} \sin^{-1}(t) + c$. Substituting $t = x^{3/2}$ back, we get: $I = \frac{2}{3} \sin^{-1}(x^{3/2}) + c$. Comparing this with the given form $\frac{2}{3} g(f(x)) + c$, we identify $f(x) = x^{3/2}$ and $g(x) = \sin^{-1}(x)$.

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

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