∫2/(1-u^2+2u)du怎么做
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/18 21:57:40
∫2/(1-u^2+2u)du怎么做∫2/(1-u^2+2u)du怎么做∫2/(1-u^2+2u)du怎么做积分:2/(1-u^2+2u)du=积分:2/[-(u^2-2u+1)+2]du=-2积分:
∫2/(1-u^2+2u)du怎么做
∫2/(1-u^2+2u)du怎么做
∫2/(1-u^2+2u)du怎么做
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))
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
∫2/(1-u^2+2u)du怎么做
求不定积分 ∫(u-1)(u^2+u+1)du
求不定积分 ∫(u-1)(u^2+u+1)du
du/(u-u^2)怎么积分
∫u/√(u^2-5)du不定积分怎么解
∫u/√(u^2-5)du不定积分怎么解
1/(u+u^2)du求不定积分
急求高手解答 积分1/(u^2-u)du怎么做
∫(u/(1+u-u^2-u^3)) du,求不定积分
=2∫[u²/(1+u)]du=2∫[(u-1)+1/(u+1)]du 这一步是怎么求出来的.
∫(u+2)/(u^2+3u)du积分
∫2u/(1-u^2)du 这积分怎么解?
∫(下限1上限1/x)[f(u)/u^2]du怎么求导
请问∫sin(u/2)*sin(u/2)du或∫sin^2 (u/2)du怎么解啊?
求一道函数积分题目∫[(1+u)/(1-2u-u^2)]du 它是怎么变成 -1/2log|u^2+2u-1|
du/(u-u^2)=dx/x怎么解,
du/（u^2-u）积分
不定积分s1/（（x-1）^1/2+1)怎么做令u=根号（x-1）得2u/1+u du 然后呢