積分: / 1/(x^2+1) dx
請俾埋步驟和let u 唔該。
數學積分開於ln?
2011-12-30 5:46 am
回答 (2)
2011-12-30 5:56 am
✔ 最佳答案
Let x = tanθ, dx = (secθ)^2 dθ∫ 1/(x^2+1) dx
= ∫ (secθ)^2/((tanθ)^2+1) dθ
= ∫ 1 dθ
= θ + C
= arctan x + C
2011-12-30 6:05 am
let x = tan u
dx/du = (sec u)^2
x^2 + 1 = (tan u)^2 + 1 = (sec u)^2
Int [1/(x^2 + 1)] dx
= Int [(sec u)^2 / (sec u)^2] du
= Int [1] du
= u + C where C is a constant
= artan x + C
dx/du = (sec u)^2
x^2 + 1 = (tan u)^2 + 1 = (sec u)^2
Int [1/(x^2 + 1)] dx
= Int [(sec u)^2 / (sec u)^2] du
= Int [1] du
= u + C where C is a constant
= artan x + C
收錄日期: 2021-04-26 19:16:33
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20111229000051KK01005