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Pure Math - Integration
2010-01-07 8:17 am
回答 (1)
2010-01-07 9:48 am
✔ 最佳答案
∫exp(-x)(1/x + 1/x^2) dx (by parts)= - exp(-x) /x - ∫exp(-x) /x^2 dx + ∫exp(-x)/x^2 dx
= - exp(-x)/x + C
so,
∫exp(-x)(1+4/x+4/x^2) dx
=∫exp(-x) dx+ 4∫exp(-x) (1/x+ 1/x^2) dx
= - exp(-x)- 4 exp(-x)/x + C
= - exp(-x) (1+4/x) + C
收錄日期: 2021-04-22 00:33:59
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100107000051KK00019