a.maths 積分題目

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

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