作者LuisSantos (^______^)
看板trans_math
標題Re: [積分]
時間Thu Feb 15 03:56:22 2007
※ 引述《king911015 (早已放棄愛上你)》之銘言：
: 1.
: ∫ln(X^2 + 2X + 2) dx =?
令 u = ln(x^2 + 2x + 2) , dv = dx
2x + 2
則 du = -------------- dx , v = x
x^2 + 2x + 2
∫ln(x^2 + 2x + 2) dx
2x + 2
= (x)(ln(x^2 + 2x + 2)) - ∫(x)(------------) dx
x^2 + 2x + 2
2x^2 + 2x
= (x)(ln(x^2 + 2x + 2)) - ∫------------ dx
x^2 + 2x + 2
2x + 4
= (x)(ln(x^2 + 2x + 2)) - ∫(2 - ------------) dx
x^2 + 2x + 2
2x + 4
= (x)(ln(x^2 + 2x + 2)) - ∫2 dx + ∫------------ dx
x^2 + 2x + 2
2x + 2 2
= (x)(ln(x^2 + 2x + 2)) - 2x + ∫------------ dx + ∫------------ dx
x^2 + 2x + 2 x^2 + 2x + 2
1
= (x)(ln(x^2 + 2x + 2)) - 2x + ln(x^2 + 2x + 2) + (2)(∫------------- dx)
(x + 1)^2 + 1
-1
= (x+1)(ln(x^2 + 2x + 2)) - 2x + (2)(tan (x + 1)) + c
: 2.
: ∫xsin^-1 xdx =?
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