∫1/(2u^2(1-u^2))du等于多少?

来源：学生作业帮助网编辑：六六作业网时间：2024/10/08 03:09:02

∫1/(2*u^2*(1-u^2))du等于多少?∫1/(2*u^2*(1-u^2))du等于多少?∫1/(2*u^2*(1-u^2))du等于多少?被积函数=1/[2u^2·(1-u^2)]=(1/

∫1/(2*u^2*(1-u^2))du等于多少?
∫1/(2*u^2*(1-u^2))du等于多少?

∫1/(2*u^2*(1-u^2))du等于多少?
被积函数 = 1/[2u^2·(1-u^2)] = (1/2)·[(1/u^2) + [1/(1 - u^2)]]
= (1/2)[u^(-2) + (1/2)[1/(1-u)] + (1/2)[1/(1+u)]]
= (1/2)u^(-2) + (1/4)[1/(1-u)] + (1/4)[1/(1+u)]
∴原积分 = (-1/2u) - (1/4)ln(1-u) + (1/4)ln(1+u) + C
=(-1/2u) + (1/4)ln[(1+u)/(1-u)] + C

求不定积分 ∫(u-1)(u^2+u+1)du 求不定积分 ∫(u-1)(u^2+u+1)du 1/(u+u^2)du求不定积分 ∫2/(1-u^2+2u)du怎么做 ∫(u/(1+u-u^2-u^3)) du,求不定积分 ∫(u+2)/(u^2+3u)du积分不定积分sin^2[u^(1/2)] du -∫ud[u/(1+u)]=-u^2/(1+u)+∫u/(1+u)du=-u^2/(1+u) + ∫du -∫1/(1+u)d(u+1) = -u^2/(1+u)+u-ln|u+1|+C求救!特别是第一步到第二步之间! 求不定积分∫du/(u-(1+u^2)^0.5/2).对不起，表述不太清楚，是∫du/(u-((1+u^2)^0.5)/2)。 =2∫[u²/(1+u)]du=2∫[(u-1)+1/(u+1)]du 这一步是怎么求出来的. 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| x(x+1)du/dx=u^2；u(1)=1 求u(x)=? ∫1/(2*u^2*(1-u^2))du等于多少? 高数:∫u/（1+2u）du=∫dx要详细步骤, 求定积分∫(2-3)u^2/(u^2-1)du 求定积分∫(1,2) 2u/(1+u) du 求不定积分.∫【 u^（1/2）+1】（u-1) du: ∫2u/(1-u^2)du 这积分怎么解?