∫(xlnx)^n 0到1的积分

∫(0->1)(xlnx)^n dx
=[1/(n+1)]∫(0->1)(lnx)^n dx^(n+1)
=[1/(n+1)] [x^(n+1).(lnx)^n ](0->1) -[n/(n+1)] ∫(0->1)x^n.(lnx)^(n-1) dx
consider
lim(x->0) x^(n+1).(lnx)^n =0
∫(0->1)(xlnx)^n dx
=-[n/(n+1)] ∫(0->1)x^n.(lnx)^(n-1) dx
=-[n/(n+1)^2] ∫(0->1)(lnx)^(n-1) dx^(n+1)
=-[n/(n+1)^2] [x^(n+1) .(lnx)^(n-1)](0->1) +[n(n-1)/(n+1)^2]∫(0->1)x^n.(lnx)^(n-2) dx
=[n(n-1)/(n+1)^2]∫(0->1)x^n.(lnx)^(n-2) dx
Inductively
∫(0->1)(xlnx)^n dx
=(-1)^n.n!/(n+1)^n]∫(0->1)x^n.dx
=(-1)^n.n!/(n+1)^(n+1)