推 Karter:+_+ 太謝謝你了,感恩~~218.168.199.230 07/08
※ 編輯: LuisSantos 來自: 61.66.173.21 (07/10 21:33)
※ 引述《Karter (偽Carter)》之銘言：
:
: (x+1)^(1/2)
: ∫----------------dx
: (x-1)^(5/2)
: 有沒有人會算,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
則 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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