[急! 20分] Integration - rewriting...

∫(x+1)/√(x^2-6x) dx [from 9 to 12]
let s=x-3
when x=9, s=6
when x=12, s=9
ds/dx=1
x+1=s+4
x^2-6x=(x-3)^2-9=s^2-9
so
∫(x+1)/√(x^2-6x) dx [from 9 to 12]
=∫(s+4)/√(s^2-9) ds [from 6 to 9]
=∫(s+4)/√(s^2-9) ds [from 6 to 9]
let u=s/3
when s=6, u=2
when s=9, u=3
du/ds=1/3
s+4=3u+4
s^2-9=9u^2-9
so
∫(s+4)/√(s^2-9) ds [from 6 to 9]
=3∫(3u+4)/√(9u^2-9) du [from 2 to 3]
=∫(3u+4)/√(u^2-1) du [from 2 to 3]
That is
∫(x+1)/√(x^2-6x) dx [from 9 to 12]
=∫(3s+4)/√(s^2-1) ds [from 2 to 3]

2007-01-05 03:12:04 補充：
若果寫得靚0的應該將上面的s用其它變數代替

好簡單咋喎=.=" 基本野.... 由於冇CHECK 數... 就咁睇~~ 覺得個ANSWER 係

a= 2 , b =3 , c = 4 , m = 3, k = 1

可能會差左3倍=.=" 你check check 啦~

2007-01-05 03:12:37 補充：
睇倒上面既博士生長篇大論=.=" 眼都突埋....為免話我抄佢.... 其實呢條數好鬼easy..你目及目及成條數最核既野係sqrt 裡面舊野。唔想佢存在既最好方法係怒屈佢係第二樣野 即係 ( x-3) /3 (你用completing square 既方法就ok 架啦!!)其他野就 就住你前者屈左既野去做就ok 架啦