Question 3 Determine an antiderivative of ∫x31−x2dx.\int \frac{x^3}{\sqrt{1-x^2}}\,dx. Show solutionHide solution+Question 3 - Solution For |x|<1|x|<1, set u=1−x2u=1-x^2, so du=−2xdxdu=-2x\,dx and x2=1−ux^2=1-u. ∫x31−x2dx=−12∫(1−u)u−1/2du=−u+13u3/2+C.\int\frac{x^3}{\sqrt{1-x^2}}\,dx=-\frac12\int(1-u)u^{-1/2}\,du =-\sqrt u+\frac13u^{3/2}+C. −1−x2+13(1−x2)3/2+C.\boxed{-\sqrt{1-x^2}+\frac13(1-x^2)^{3/2}+C.} Differentiation gives x/1−x2−x1−x2=x3/1−x2x/\sqrt{1-x^2}-x\sqrt{1-x^2}=x^3/\sqrt{1-x^2}.
