问题描述: 求有理函数不定积分 x/x^2-2x+2 dx 1个回答 分类:数学 2014-09-17 问题解答: 我来补答 ∫x/(x^2-2x+2) dx=∫(x-1+1)/(x^2-2x+2) dx=∫(x-1)/(x^2-2x+2) dx+∫1/(x^2-2x+2) dx=1/2∫1/(x^2-2x+2) d(x^2-2x+2)+∫1/[(x-1)^2+1] dx=1/2ln(x^2-2x+2)+arctan(x-1)+C 展开全文阅读