∫(1-x)/[√(9-4x^2)]dx=

来源：学生作业帮助网编辑：六六作业网时间：2024/12/23 00:01:06

∫(1-x)/[√(9-4x^2)]dx=∫(1-x)/[√(9-4x^2)]dx=∫(1-x)/[√(9-4x^2)]dx=∫(1-x)/[√(9-4x^2)]dx=5-离问题结束还有14天23小时

∫(1-x)/[√(9-4x^2)]dx=
∫(1-x)/[√(9-4x^2)]dx=

∫(1-x)/[√(9-4x^2)]dx=
∫(1-x)/[√(9-4x^2)]dx=
5 - 离问题结束还有 14 天 23 小时
数学
kkcl111 - 试用期一级
回答
换元法令x=3/2sint,t∈[-0.5π,0.5π]
带入后得到
∫(1-x)/[√(9-4x^2)]dx=∫(1-1.5sint)1.5costdt/3cost
=∫(1-1.5sint)0.5dt
=0.5t+0.75cost+C=0.5arcsin2/3x+1/4√9-4x^2+C

∫(1-x)/[√(9-4x^2)]dx
=
∫1/[√(9-4x^2)]dx+∫x/[√(9-4x^2)]dx
=
arcsin(2x/3)/2-1/4*[√(9-4x^2)]+c
这儿用到积分公式
1）∫1/根号下(a^2-x^2)dx=arcsin(x/a)+c

我只公布标准答案，供答题者参考，楼主不要给分
Sqrt[9 - 4x^2]/4 + ArcSin[(2x)/3]/2+C

∫(1-x)dx/(9-4x^2)^(1/2)
= ∫dx/(9-4x^2)^(1/2) - ∫xdx/(9-4x^2)^(1/2)
= (1/2)∫d(2x/3)/[1-(2x/3)^2]^(1/2) + (1/8)∫d(9-4x^2)/(9-4x^2)^(1/2)
= (1/2)arcsin(2x/3) + (1/4)(9-4x^2)^(1/2) + C.
C = const.

∫(1-x)/[√(9-4x^2)]dx= ∫1-x/(√9-4x^2)dx, ∫(x-1)/√(9-4x^2)dx不定积分求不定积分 ∫ 1/(1+2x)² dx ∫ x/√x²+4 dx ∫1/√x*(4-x)dx ∫dx/[x√(1-x^4)] ∫x√(1+2x)dx ∫x/(4+9x^2)dx ∫（x^2+1/x^4)dx ∫1/(x^4-x^2)dx 积分符号∫ 2^x/(1+4^x) dx积分符号∫(X+1)/(x^2+9) dx积分符号∫(tg根号X) / 根号X dx x-9/[(根号)x]+3 dx ∫ x+1/[(根号)x] dx ∫ [（3-x^2）]^2 dx ①∫（1-x）/√9-4x² dx②∫√x²-9/x dx③∫dx/√1+e^x 不定积分∫dx(1-x)/√(9-4x) 求不定积分∫x/(√4-9x^2)dx 求不定积分∫1/x√(x^2-9)dx ∫[-1,1](x+√4-x^2)^2dx 不定积分∫(x-1)/√(1-4x^2)dx