∫2/(1-u^2+2u)du怎么做
问题描述：
∫2/(1-u^2+2u)du怎么做
答
1-u^2+2u＝2-(u-1)^2＝(√2)^2-(u-1)^2，du＝d(u-1)，所以
∫2/(1-u^2+2u)du＝∫2/[(√2)^2-(u-1)^2]d(u-1)，套用不定积分公式∫1/(a^2-x^2)dx＝1/(2a)×ln|(a+x)/(a-x)|＋C得
∫2/(1-u^2+2u)du＝1/√2×ln|(√2+u-1)/(√2-u+1)|＋C
答
积分:2/(1-u^2+2u)du
=积分:2/[-(u^2-2u+1)+2]du
=-2积分:1/[(u-1)^2-(根号2)^2]du
=-2积分:1/[(u-1)^2-(根号2)^2]d(u-1)
=-2/ln2*ln|(u-1-根号2)/(u-1+根号2)|+C
(C为常数)
有公式:
积分:dx/(x^2-a^2)
=1/2aln|(x-a)/(x+a)|+C
提示:
将:1/(x^2-a^2)=1/2a*(1/(x-a)-1/(x+a))