记 仔蔽此f(x) = ∑(n≥1)n[(x-1)^(n-1)],积分,得 ∫[0,x]f(t)dt = ∑并春(n≥1)[(x-1)^n] = -1+∑(n≥0)[(x-1)^n] = -1+1/[1-(x-1)] ,|x-1|<1 =-1+1/(2-x),0 求导,得 f(x) = [-1+1/(2-x)]' 念迅 = 1/(2- x)²,0于是, g.e. = (x-1)*f(x) = ……。