∫x^3/√(1-x^2)dx=∫x^2*x/√(1-x^2)dx=1/2∫x^2/√(1-x^2)dx^2；令√(1-x^2)=t,则x^2=1-t^2,dx^2=d(1-t^2)=-2tdt,则原式可化为∫(t^2-1)dt=1/3t^3-t+C=1/3（1-x^2)^(3/2)-√(1-x^2)+C