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If $\int \sqrt{x}\left(1-x^3\right)^{\frac{-1}{2}} d x=\frac{2}{3} g(f(x))+c$, then
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$f(x)=x^{\frac{3}{2}}, g(x)=\sin ^{-1} x$
$I=\int \sqrt{x}\left(1-x^3\right)^{-\frac{1}{2}} d x$
Let $\mathrm{x}^{\frac{3}{2}}=\mathrm{t} \Rightarrow \frac{3}{2} \mathrm{x}^{\frac{1}{2}} \mathrm{dx}=\mathrm{dt} \Rightarrow \sqrt{\mathrm{x}} \mathrm{dx}=\frac{2}{3} \mathrm{dt}$ $\therefore \mathrm{I}=\int\left(1-\mathrm{t}^2\right)^{-\frac{1}{2}} \frac{2}{3} \mathrm{dt}=\frac{2}{3} \int \frac{1}{\sqrt{1-\mathrm{t}^2}} \mathrm{dt}$
$\begin{aligned} & I=\frac{2}{3} \sin ^{-1} t+c \\ & I=\frac{2}{3} \sin ^{-1}\left(x^{\frac{3}{2}}\right)+c\end{aligned}$
Let $f(x)=x^{3 / 2} \& g(x)=\sin ^{-1} x$
Then, $I=\frac{2}{3} g(f(x))+c$
Let $\mathrm{x}^{\frac{3}{2}}=\mathrm{t} \Rightarrow \frac{3}{2} \mathrm{x}^{\frac{1}{2}} \mathrm{dx}=\mathrm{dt} \Rightarrow \sqrt{\mathrm{x}} \mathrm{dx}=\frac{2}{3} \mathrm{dt}$ $\therefore \mathrm{I}=\int\left(1-\mathrm{t}^2\right)^{-\frac{1}{2}} \frac{2}{3} \mathrm{dt}=\frac{2}{3} \int \frac{1}{\sqrt{1-\mathrm{t}^2}} \mathrm{dt}$
$\begin{aligned} & I=\frac{2}{3} \sin ^{-1} t+c \\ & I=\frac{2}{3} \sin ^{-1}\left(x^{\frac{3}{2}}\right)+c\end{aligned}$
Let $f(x)=x^{3 / 2} \& g(x)=\sin ^{-1} x$
Then, $I=\frac{2}{3} g(f(x))+c$
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