先做换元u=x^2，1/2∫[(1-u^2)]^(3/2)du，然后再用三角换元
∫xdx/√(1-x^4)^3=(1/2)∫dx^2/√(1-x^4)^3x^2=sinu=(1/2)∫dsinu/cosu^3=(1/2)∫du/cosu^2=(1/2)tanu+C=(1/2)x^2/√(1-x^4) +C
