Integration
回答 (1)
2010-05-20 4:48 pm
✔ 最佳答案
By partial fractions,2/(x^2 - 1) = 1/(x - 1) - 1/(x + 1)
So, S 2/(x^2 - 1) dx
= S [1/(x - 1) - 1/(x + 1)]dx
= S 1/(x - 1) dx - S 1/(x + 1) dx
= ln(x - 1) - ln(x + 1) + C
= ln[(x - 1)/(x + 1)] + C
where C is a constant
參考: Prof. Physics
收錄日期: 2021-04-19 22:19:15
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100520000051KK00200