Provide Integrate 1/(y^2 -1 ) = 1/2 In[(y-1)/(y+1)]
Step plx
Integration
2012-09-10 6:21 am
回答 (2)
2012-09-10 7:00 am
✔ 最佳答案
圖片參考：http://i1096.photobucket.com/albums/g340/pingshek/2012-09-09.png?t=1347202770
http://i1096.photobucket.com/albums/g340/pingshek/2012-09-09.png?t=1347202770
參考: miraco
2012-09-10 6:47 am
∫1/(y^2 -1 ) dy
=∫1/[(y+1)(y-1)] dy
=∫1/2(y-1)dy-∫1/2(y+1)dy
=(1/2)ln(y-1)-(1/2)ln(y+1)+C
=(1/2)ln[(y-1)/(y+1)]+C
=∫1/[(y+1)(y-1)] dy
=∫1/2(y-1)dy-∫1/2(y+1)dy
=(1/2)ln(y-1)-(1/2)ln(y+1)+C
=(1/2)ln[(y-1)/(y+1)]+C
收錄日期: 2021-04-13 18:57:49
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120909000051KK00829