作者Fubini (===漂移的阿尼===)
看板trans_math
標題Re: [積分] 關於不定積分...
時間Thu Mar 27 18:51:24 2008
※ 引述《jeremyhcw (( ￣ c￣)y▂ξ)》之銘言：
: 第一名改正為
: ∫1/x(5x-6-x^2)^(1/2)dx
dx
∫------------------
x(5x-6-x^2)^(1/2)
dx
= ∫-------------------- 令y=x-2 , x=y+2 , dx=dy
x [(3-x)(x-2)]^(1/2)
dy
= ∫-------------------- 分子分母同除y
(y+2) (y-y^2)^(1/2)
dy
= ∫------------------------ 令t=1/y , y=1/t , dy= - (1/t^2)dt
y (y+2) [(1/y)-y)^(1/2)
-t
= ∫------------------------ dt
(t^2)[(1/t)+2](t-1)^(1/2)
1
= -∫--------------------- dt 令u=(t-1)^(1/2) , t=u^2 +1 , dt=2udu
t[(1/t)+2](t-1)^(1/2)
du -1
= -2∫----------- = - (2/3)^(1/2) tan ((2/3)^(1/2) * u + c
2u^2+3
-1
= - (2/3)^(1/2) tan ((2(t-1)/3)^(1/2) + c
-1
= - (2/3)^(1/2) tan {[2(1/y-1)]/3}^(1/2) + c
-1 6-2x
= - (2/3)^(1/2) tan √(------) + c
3x-6
隨便算的 答案僅供參考.
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