求导f'(x)=∑（-1）^n×x^(2n-1)=-x×∑（-x^2）^(n-1)=-x×1/(1+x^2)=-x/(1+x^2).积分,f(x)=f(0)+∫(0到x) f'(t)dt=1-1/2×ln(1+x^2).令f'(x)=0得x=0,x＞0时f'(x)＜0；x＜0时f'(x)＞0,所以f(0)是f(x)的极大值,极大...