连续用凑微分法即可
∫(1,0) x^m*(1-x)^n dx
=∫(1,0) (1-x)^n / (m+1)d(x^m+1)
=(1-x)^n/(m+1)*x^(m+1) - ∫(1,0) n(1-x)^(n-1) * (-1) * x^(m+1)dx
前面的一项显然为0
=n/(m+1)∫(1,0) (1-x)^(n-1) * x^(m+1)dx
继续用凑微分,直到1-x的次数为0
=n!/[(m+1)(m+2)...(m+n)]∫(1,0)x^(m+n)dx
=n!/[(m+1)(m+2)...(m+n)(m+n+1)]
=(n!m!)/(m+n+1)!
（完成）(1-x)^n/(m+1)*x^(m+1)这项为什么为0？F(x)=(1-x)^n/(m+1)*x^(m+1)所以F(1)=0,F(0)=0F(1)-F(0)=0