F.6 數學定積分問題 10
回答 (2)
2009-10-20 8:05 am
✔ 最佳答案
∫1/[1 + e^(2x)]dxLet 1 + e^(2x) = u
2e^(2x) dx = du
2(u - 1) dx = du
dx = du /[2(u - 1)]
∫1/[1 + e^(2x)]dx
= ∫(1/u)]/[2(u - 1)] du
= (1/2)∫1/[u(u -1)] du
= (1/2)∫[1/(u - 1) - 1/u]du
= (1/2)[ln(u - 1) - ln u] + C
= (1/2)[ln(e^2x) - ln(1 + e^2x)] + C
= x - (1/2)ln(1 + e^2x) + C
2009-10-22 7:15 am
= (1/2)∫1/[u(u -1)] du
= (1/2)∫[1/(u - 1) - 1/u]du
請問呢一步係點變家?可唔可以詳細d教下我喔?麻煩晒!~
= (1/2)∫[1/(u - 1) - 1/u]du
請問呢一步係點變家?可唔可以詳細d教下我喔?麻煩晒!~
收錄日期: 2021-04-23 23:18:29
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091019000051KK01806