http://www.beacon.com.hk/mockxchange/Amaths/AmathII/AmathIIqa.htm
第1題
a.maths 積分題目
2007-04-16 12:45 am
回答 (2)
2007-04-16 3:18 am
✔ 最佳答案
我用$ 黎代表 積分符號..因為唔知點解唔可以將math equator d 野貼上黎...
里條數應該都out 左c...因為要用到changing variable..
咁我都幫你解下...
$(x*(x^2+1)^-5 dx
=1/2*$(x^2+1)^-5 2xdx
=1/2*$(x^2+1)^-5 d(x^2+1)
=1/2*$u^-5 du let u=x^2+1
=1/2*(u^-4)/-4+C
=-1/8 u^-4
=1/(8*(x^2+1)^4)+C
2007-04-16 12:51 am
In x (x2 +1)-5 dx
= In 1/2 * (x2 +1)-5 (2xdx)
= In (x2 +1)-5 (d(x2 + 1))
(you can just treat it as In u-5 du)
= (-1/4) * (x2 +1)-4 + C
= -1/(4(x2 +1)4) + C, where C is a constant
2007-04-16 04:02:23 補充：
sorry i missed the 1/2 at the third lineso it should be -1/8 instead
2007-04-16 19:24:39 補充：
http://www.hkeaa.edu.hk/doc/sd/2007ce%28e%29%5Fam%2Epdfintegration by substitution is NOT required
收錄日期: 2021-04-12 22:24:54
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070415000051KK03666