题目

积分∫arctanx*x^2/(1+x^2)dx
应该用分部积分法,可我怎么算也不和答案上一样

提问时间：2020-10-28

答案

∫[arctan(x)]*x^2/(1+x^2)dx = ∫1*arctanxdx-∫(arctanx)/(x^2+1)dx = {x*arctan(x)-∫x/(x^2+1)dx}-∫[arctan(x)]d[arctan(x)] = x*arctan(x)-{ln(x^2+1)+[arctan(x)]^2}/2+C .