求不定积分∫(1/x^2+2x+5)dx

结果为：(1/2)arctan[(x+1)/2]+ C 解题过程如下： 原式=∫1/(x^2+2x+5)dx =∫1/[(x+1)^2+4]dx =∫(1/4)/[ [(x+1)/2]^2+1]dx =∫(1/4)?2/[ [(x+1)/2]^2+1]d( (x+1)/2) =(1/2)∫1/[ [(x+1)/2]^2+1]d( (x+1)/2) =(1/2)arctan[(x+