[x(1-x^2)^(3/2)]/3dx在[1,0]的积分怎么求的

(1/3)∫[0---->1] x(1-x²)^(3/2)dx令1-x²=u²,两边微分得：xdx=-udu,u：1--->0,(1-x²)^(3/2)=u³,=-(1/3)∫[1---->0] u³udu=(1/3)∫[0---->1] u⁴du=(1/3)(1/5)u⁵ |[0---->1...

最讨厌积分了

具体见图片