问题描述: ∫(x-1)/(x^2+2x+3)dx的不定积分怎么求 1个回答 分类:数学 2014-10-29 问题解答: 我来补答 ∫(x-1)/(x²+2x+3)dx=½∫(2x-2)/(x²+2x+3)dx=½∫(2x+2-4)/(x²+2x+3)dx=½∫(2x+2)/(x²+2x+3)dx - ½∫4/(x²+2x+3)dx=½∫(2x+2)/(x²+2x+3)dx - 2∫1/(x²+2x+3)dx=½∫d(x²+2x+3)/(x²+2x+3) - 2∫1/[(x+1)²+2]dx=½ln|x²+2x+3| - ∫1/{[(x+1)/√2]²+1}dx + C=½ln|x²+2x+3| - (√2)∫1/{[(x+1)/√2]²+1}d[(x+1)/√2] + C=½ln|x²+2x+3| - (√2)arctan[(x+1)/√2] + C 展开全文阅读