x2⼀(1+x2)dx的不定积分怎么求

2025-02-25 19:11:51
推荐回答(1个)
回答1:

∫x^2/(1+x^2)dx
=∫(x^2+1-1)/(1+x^2)dx
=∫(1-1/(1+x^2))dx
=x-arctanx+C