∫(1-x)/[√(9-4x^2)]dx=

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∫(1-x)/[√(9-4x^2)]dx=
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∫(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.

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