问题描述: 求不定积分∫x/(x^2+2x+2)dx 1个回答 分类:数学 2014-11-07 问题解答: 我来补答 解∫x/(x²+2x+2)dx=1/2∫(2x+2-2)/(x²+2x+2)dx=1/2∫(2x+2)/(x²+2x+2)dx-∫1/(x²+2x+2)dx=1/2∫1/(x²+2x+2)d(x²+2x+2)-∫1/[(x+1)²+1]dx=1/2∫1/udu-∫1/[(x+1)²+1]d(x+1)=1/2ln|u|-∫1/(u²+1)du=1/2ln(x²+2x+2)-acrtanu+C=1/2ln(x²+2x+2)-arctan(x+1)+C 展开全文阅读