问题描述: 积分∫x/[(x^2+1)(x^2+4)]dx 1个回答 分类:数学 2014-11-26 问题解答: 我来补答 ∫x/[(x^2+1)(x^2+4)]dx=1/3∫x[1/(x^2+1) - 1/(x^2+4)]dx=1/3[∫x/(x^2+1)dx-∫x/(x^2+4)dx]=1/3[1/2∫1/[(x^2+1)]d(x^2+1)-1/2∫1/[(x^2+4)d(x^2+4)]]=1/3[1/2ln(x^2+1)-1/2ln(x^2+4)+C]=1/6ln[(x^2+1)/(x^2+4)]+C 展开全文阅读