I don't know
∫ x / ( 1+x^2 ) = ?
Integration F.6
2008-07-03 6:50 am
回答 (3)
2008-07-03 7:21 am
✔ 最佳答案
∫x/(1+x)dx=(1/2)∫1/(1+x)d(1+x)
=(1/2)ln|1+x|+C, where C is a constant
2008-07-02 23:22:48 補充：
yahoo出唔到某D字元，我打多次
∫x/(1+x²)dx
=(1/2)∫1/(1+x²)d(1+x²)
=(1/2)ln|1+x²|+C, where C is a constant
2008-07-03 5:34 pm
Another substitution method:
http://i117.photobucket.com/albums/o61/billy_hywung/Jul08/Crazyint1.jpg
http://i117.photobucket.com/albums/o61/billy_hywung/Jul08/Crazyint1.jpg
2008-07-03 7:39 am
let x=tanθ
∫ x / ( 1+x^2 ) dx
=∫ tanθ / ( sec^2θ ) (sec^2θ) dθ
=∫ tanθ dθ
=∫-1/cosθ d (cosθ)
=-ln(cosθ)+C
=-ln(1/√(1+x^2))+C
=ln(√(1+x^2))+C
∫ x / ( 1+x^2 ) dx
=∫ tanθ / ( sec^2θ ) (sec^2θ) dθ
=∫ tanθ dθ
=∫-1/cosθ d (cosθ)
=-ln(cosθ)+C
=-ln(1/√(1+x^2))+C
=ln(√(1+x^2))+C
收錄日期: 2021-04-26 13:45:09
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080702000051KK03265