求不定积分∫(1/x^2+2x+5)dx

题目

求不定积分∫(1/x^2+2x+5)dx
有搜到答案是这样的,但是不懂怎么从倒数第二步到最后一步
∫1/(x^2+2x+5)dx
=∫1/[(x+1)^2+4]dx
=∫1/[(x+1)^2+2^2]d(x+1)
=(1/2)arctan[(x+1)/2]+C

答案

∫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+1)/2]+ C上面对你搜到的答案进行了细化.主要还是利...