∫ z/(1-4z^2)dz = - 1/8 log l1-4z^2l + C
I do not know how to find -1/8 in this answer. I am confused. Please help me if you know?
Thanks!
Can you help me with INDEFINITE INTEGRAL?
2016-06-20 10:09 pm
回答 (1)
2016-06-20 10:16 pm
✔ 最佳答案
Let u = 1 - 4z²Then, du = -8z dz
∫ [z/(1 - 4z²)] dz
= ∫ [1/(1 - 4z²)] z dz
= ∫ [1/(1 - 4z²)] [-(1/8)] (-8z dz)
= -(1/8) ∫ [1/(1 - 4z²)] (-8z dz)
= -(1/8) ∫ [1/u] du
= -(1/8) ln|u| + C
= -(1/8) ln|1 - 4z²| + C
收錄日期: 2021-04-18 15:07:48
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160620140954AAtGKff