设f(x)=(x-1)ιn(x-1)+x,则f(ex+1)=()。
A.xex+ex+1
B.xex+ex+1
C.xex+ex
D.xex
设函数f(x)在[0,1]上有二阶连续导数,且f(0)=f(1)=0,f(x)≠0,x∈(0,1),证明∫(1,0)f(x)dx=1/2∫(1,0)x(x-1)f"(x)dx
若f(x-1)=x2(x-1),则f(x)=[ ].
A.x(x+1)2
B.x(x-1)2
C.x2(x+1)
D.x2(x-1)
A.[0,1]
B.[-1/4,3/4]
C.[1/4,3/4]
D.[-1/4,5/4]