What is \(\displaystyle\int \dfrac{\sqrt{x}}{\sqrt{1-x^3}}\,dx\) equal to?
\(\dfrac23\sin^{-1}(x^{3/2})+c\)
Substitute \(u=x^{3/2}\), so \(du=\dfrac32\sqrt{x}\,dx\), that is, \(\sqrt{x}\,dx=\dfrac23\,du\). Also \(x^3=u^2\), so the integral becomes \(\displaystyle\int\dfrac{2}{3}\dfrac{du}{\sqrt{1-u^2}}=\dfrac23\sin^{-1}(u)+c=\dfrac23\sin^{-1}(x^{3/2})+c\).
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The value of \(\rm \int \frac{(x+1)}{x(xe^x + 1)}\ dx\) is equal to:
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