x^2+2x+5 = (x+1)^2 +4
let
x+1 = 2tanu
dx=2(secu)^2 du
∫x/(x^2+2x+5) dx
=(1/2) ∫(2x+2)/(x^2+2x+5) dx -∫dx/(x^2+2x+5)
=(1/2)ln|x^2+2x+5| -∫dx/(x^2+2x+5)
=(1/2)ln|x^2+2x+5| -∫2(secu)^2 /[4(secu)^2 ] du
=(1/2)ln|x^2+2x+5| -(1/2)u +C
=(1/2)ln|x^2+2x+5| -(1/2)arctan[(x+1)/2] + C
内容全部来源于网络收集，如有侵权，请联系网站删除：QQ：24596024