∫01x(1−x)ndx=
1n+1
1n+2
1(n+1)(n+2)
2n+3(n+1)(n+2)
∫01x(1−x)ndx,x→1−x
=∫01(1−x)xndx=∫01(xn−xn+1)dx
=1n+1−1n+2=1(n+1)(n+2)