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Let $\int \frac{x^{1 / 2}}{\sqrt{1-x^3}} d x=\frac{2}{3} g(f(x))+c$; then
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Verified Answer
The correct answer is:
$f(x)=x^{3 / 2}, g(x)=\sin ^{-1} x$
$\int \frac{x^{1 / 2} d x}{\sqrt{1-\left(x^{3 / 2}\right)^2}}=2 / 3 \int \frac{d t}{\sqrt{1-t^2}}$
$x^{3 / 2}=t$
$3 / 2 x^{1 / 2} d x=d t$
$=2 / \sin ^{-1}(t)+c$
$=2 / 3 \sin ^{-1}\left(x^{3 / 2}\right)+c$
$x^{3 / 2}=t$
$3 / 2 x^{1 / 2} d x=d t$
$=2 / \sin ^{-1}(t)+c$
$=2 / 3 \sin ^{-1}\left(x^{3 / 2}\right)+c$
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