记 f(x) = ∑(n>=1){[(-1)^(n-1)](x^n)/n},-1求导,得 f'(x) = ∑(n>=1){[(-1)^(n-1)][x^(n-1)]} = 1/(1+x),-1积分,得 f(x) = ∫[0,x][1/(1+t)]dt = ln(1+x),-1