大一简单数学题，求不定积分？
展开全部
∫(x-2)/(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| - (1/2)arctan[(x-1)/2] +C
//
consider
x^2-2x+5 = (x-1)^2 +4
let
x-1=2tanu
dx =2(secu)^2 du
∫dx/(x^2-2x+5)
=∫2(secu)^2 du/[4(secu)^2 ]
=(1/2)u + C
=(1/2)arctan[(x-1)/2] +C
=(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| - (1/2)arctan[(x-1)/2] +C
//
consider
x^2-2x+5 = (x-1)^2 +4
let
x-1=2tanu
dx =2(secu)^2 du
∫dx/(x^2-2x+5)
=∫2(secu)^2 du/[4(secu)^2 ]
=(1/2)u + C
=(1/2)arctan[(x-1)/2] +C
本回答被提问者采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
