作者LuisSantos (^______^)
看板trans_math
标题Re: 积分
时间Sat Feb 17 12:39:41 2007
※ 引述《king911015 (早已放弃爱上你)》之铭言：
: １.
: 1
: ∫ ln (X-1) dx =?
: 0
先计算 ∫ln(x-1) dx
令 u = ln(x-1) , dv = dx
1
则 du = ----- dx , v = x
x - 1
∫ln(x-1) dx
1
= (x)(ln(x-1)) - ∫(x)(-----) dx
x - 1
1
= (x)(ln(x-1)) - ∫(1 + -----) dx
x - 1
= (x)(ln(x-1)) - (x + ln(x-1)) + c
= (x)(ln(x-1)) - x - ln(x-1) + c
= (x-1)(ln(x-1)) - x + c
1
∫ ln(x-1) dx
0
1/2 1
= ∫ ln(x-1) dx + ∫ ln(x-1) dx
0 1/2
1/2 b
= lim ∫ ln(x-1) dx + lim ∫ ln(x-1) dx
a→0 a b→1- 1/2
|1/2 |b
= lim (x)(ln(x-1)) - x | + lim (x)(ln(x-1)) - (x) |
a→0 |a b→1- |1/2
= 不存在
: 2.
: 3 1
: ∫ ------------- dx =?
: 2 x(x + 3)^2
: 3.
: √3／2
: X^3
: ∫ --------------- dx = ?
: 0 √(1 - X^2)
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