Help me solve this hard question,please.
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Indefinite integration
2011-11-19 2:50 am
回答 (3)
2011-11-19 6:42 am
✔ 最佳答案
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參考: 原創答案
2011-11-19 6:52 am
let u^3=3x+2
∴x=1/3(u^3-2)
∴dx=u^2du
∫1/[x-∛(3x+2)]dx
=∫=∫(3u^2)/(u^3-3u-2)du
=∫(3u^2-3+3)/(u^3-3u-2)du
=∫(3u^2-3)/(u^3-3u-2)du+∫3/(u^3-3u-2)du
=∫1/(u^3-3u-2)d(u^3-3u-2)+∫{[(1/3)/(u-2)]+[(-1/3)/(u+1)]+[(-1)/(u+1)^2]}du
=ln(u^3-3u-2)+ln(u-2)/3-ln(u+1)/3+1/(u+1)+C
=後面自己化反一起同將u變反x甘就得噶拉...因為打出來太麻煩啦...
∴x=1/3(u^3-2)
∴dx=u^2du
∫1/[x-∛(3x+2)]dx
=∫=∫(3u^2)/(u^3-3u-2)du
=∫(3u^2-3+3)/(u^3-3u-2)du
=∫(3u^2-3)/(u^3-3u-2)du+∫3/(u^3-3u-2)du
=∫1/(u^3-3u-2)d(u^3-3u-2)+∫{[(1/3)/(u-2)]+[(-1/3)/(u+1)]+[(-1)/(u+1)^2]}du
=ln(u^3-3u-2)+ln(u-2)/3-ln(u+1)/3+1/(u+1)+C
=後面自己化反一起同將u變反x甘就得噶拉...因為打出來太麻煩啦...
2011-11-19 3:45 am
Sub. u³= 3x+2
收錄日期: 2021-04-16 13:44:08
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20111118000051KK00601