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将函数(arctanx^2)⼀x展开成x的幂级数

2025-03-14 20:57:51

推荐回答（1个）

回答1：

y=arctanx
求导y'=1/(1+x^2)=1-x^2+x^4-x^6+...
积分还原：y=x-x^3/3+x^5/5-x^7/7+....
因此有：
arctanx^2=x^2-x^6/3+x^10/5-x^14/7+....
(arctanx^2)/x=x-x^5/3+x^9/5-x^13/7+....-(-1)^nx^(4n-3)/(2n-1)+..

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