你说的是:
(1)F(x)=∫[0,x]xf(t)dt=x∫[0,x]f(t)dt,
F'(x)=∫[0,x]f(t)dt+xf(x).
(2)G(x)=∫[0,x]tf(2x-t+1)dt,
先做变量替换u=2x-t+1,则t=2x-u+1,dt=-du
G(x)=∫[0,x]tf(2x-t+1)dt
=∫[2x+1,x+1](2x-u+1)f(u)(-1)du
=∫[x+1,2x+1](2x-u+1)f(u)du
=2x∫[x+1,2x+1]f(u)du-∫[x+1,2x+1]uf(u)du+∫[x+1,2x+1]f(u)du,
于是,
G'(x)=(d/dx){2x∫[x+1,2x+1]f(u)du-∫[x+1,2x+1]uf(u)du+∫[x+1,2x+1]f(u)du}
=……
(用求导法则和积分上限函数的求导法来求解,这里不好写,留给你了)