作者LuisSantos (^______^)
看板trans_math
标题Re: 不定积分
时间Fri Jul 8 00:34:28 2005
※ 引述《Karter (伪Carter)》之铭言：
:
: (x+1)^(1/2)
: ∫----------------dx
: (x-1)^(5/2)
: 有没有人会算,感恩~~~ <(_ _)>
(x+1)^(1/2)
I = ∫----------------dx
(x-1)^(5/2)
(x+1)^(1/2) 1
= ∫-------------- ---------- dx
(x-1)^(1/2) (x-1)^2
(x+1)^(1/2) x+1 -2
令 u = -------------- => ----- = u^2 => ---------dx = 2udu
(x-1)^(1/2) x-1 (x-1)^2
(x+1)^(1/2) 1
则 I = ∫---------------- ---------- dx
(x-1)^(1/2) (x-1)^2
= ∫u*(-udu)
= (-1)*∫u^2 du
1
= - ---u^3 + c
3
1 (x+1)^(3/2)
= - --- ------------- + c
3 (x-1)^(3/2)
--
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◆ From: 61.66.173.21
1F：推 Karter:+_+ 太谢谢你了,感恩~~218.168.199.230 07/08
※ 编辑: LuisSantos 来自: 61.66.173.21 (07/10 21:33)