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Find f[(b^2+6b+4)/(b+1)]db
我想問關於1條Integral問題點計
2009-06-02 8:06 am
回答 (2)
2009-06-02 8:11 am
✔ 最佳答案
As follows:圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/22-1.gif
2009-06-02 00:14:28 補充：
Note that ∫(1/x)dx=ln│x│+c
2009-06-02 6:41 pm
Let b + 1 = u
so db = du
b^2 + 6b + 4 = (b + 1)^2 + 4(b + 1) - 1 = u^2 + 4u - 1.
so the integral becomes
S (u^2 + 4u - 1)du/u = S (u + 4 - 1/u)du
= u^2 + 4u - lu u + C
= (b + 1)^2 + 4(b + 1) - ln ( b + 1) + C.
2009-06-02 10:45:49 補充：
Correction : integrating u should be u^2/2, not u^2, so answer should be
(b + 1)^2/2 + 4(b + 1) - ln (b + 1) + C.
so db = du
b^2 + 6b + 4 = (b + 1)^2 + 4(b + 1) - 1 = u^2 + 4u - 1.
so the integral becomes
S (u^2 + 4u - 1)du/u = S (u + 4 - 1/u)du
= u^2 + 4u - lu u + C
= (b + 1)^2 + 4(b + 1) - ln ( b + 1) + C.
2009-06-02 10:45:49 補充：
Correction : integrating u should be u^2/2, not u^2, so answer should be
(b + 1)^2/2 + 4(b + 1) - ln (b + 1) + C.
收錄日期: 2021-04-22 00:38:48
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