integral of from 1 to 2
(4y^2-7y^2-12)/[y(y+2)(y-3)dx
(4y^2-7y^2-12)/[y(y+2)(y-3)=a/y+b/(y+2)+c/(y+3)
(4y^2-7y^2-12)=a(y+2)(y-3)+by(y-3)+cy(y+2)
y=0>>=-12=-6a>>a=2
y=-2>>b=5/9
y=3>>c=1/5
2/y+5/9(1/y+2)+1/5(1/y-3)
from 1 to 2
2lny+1/9ln(y+2)+1/5ln(y-3)
我計出黎好似有唔岩...都唔知怎解
ans:(27/5)ln2-(9/5)ln3.....or (9/5)ln2/3
partial fraction
2008-02-14 9:09 pm
回答 (2)
2008-02-14 9:50 pm
✔ 最佳答案
Solution:圖片參考：http://hk.geocities.com/stevieg_1023/II1.gif
2008-02-14 13:50:44 補充：
問題係你個b,唔係5/9,係9/5~!!!
2008-02-14 10:07 pm
from your answer:
integral of from 1 to 2
(4y^2-7y^2-12)/[y(y+2)(y-3)dx
(4y^2-7y^2-12)/[y(y+2)(y-3)=a/y+b/(y+2)+c/(y+3)
(4y^2-7y^2-12)=a(y+2)(y-3)+by(y-3)+cy(y+2)
put y=0
>>-12=-6a
>>a=2 (correct)
put y=-2
>> -24 = -10b
>>b=5/9 (incorrect)
>>b = 12/5
put y=3
>> -39 = 15c
>>c=1/5 (incorrect)
>> c = -13/5
>a/y+b/(y+2)+c/(y+3)
>2/y+[12/5 / (y+2)] - [13/5 / (y+3)]
........
integral of from 1 to 2
(4y^2-7y^2-12)/[y(y+2)(y-3)dx
(4y^2-7y^2-12)/[y(y+2)(y-3)=a/y+b/(y+2)+c/(y+3)
(4y^2-7y^2-12)=a(y+2)(y-3)+by(y-3)+cy(y+2)
put y=0
>>-12=-6a
>>a=2 (correct)
put y=-2
>> -24 = -10b
>>b=5/9 (incorrect)
>>b = 12/5
put y=3
>> -39 = 15c
>>c=1/5 (incorrect)
>> c = -13/5
>a/y+b/(y+2)+c/(y+3)
>2/y+[12/5 / (y+2)] - [13/5 / (y+3)]
........
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