简单计算一下即可，答案如图所示
1.s(x)=∑(∞,n=0)x^(2n+1)/(2n+1),s'(x)=∑(∞,n=0)x^2n=1/(1-x^2)收敛域为（-1，1）2.s(x)=∫(0,x)1/(1-x^2)dx=1/2∫(0,x)[1/(1+x)+1/(1-x)]dx=1/2ln(1+x)/(1-x)x取值范围（-1,1)
