求 1/(x^2-x+1) 的不定積分，想請問過程，該怎麼算？
回答 (3)
2015-10-13 12:20 am
✔ 最佳答案
阿災的回答是微分, 不是積分...YOUTUBE上有人做這個的詳解, 可參考看看
https://www.youtube.com/watch?v=ORXOLN5HQDI
答案是(2/√3)*[tan^(-1)[(2x-1)/√3] + C,
tan^(-1)=arctan
2015-10-13 12:47 am
不定積分∫dx/(x^2-x+1) = ?
= ∫dx/(x^2 - x + 1/4 + 3/4)
= ∫dx/[(x-1/2)^2 + 3/4]
= ∫d(x-1/2)/[(x-1/2)^2 + 3/4]
= ∫dy/(y^2 + 3/4) ;;; y = x - 1/2
= ∫dy/(y^2 + a^2) ;;; a = √3/2
= (1/a)∫d(y/a)/[(y/a)^2 + 1]
= (1/a)∫dz/(z^2 + 1) ;;; z = y/a
= [atan(z)]/a + c
= [atan(y/a)]*2/√3 + c
= [atan(2y/√3)]*2/√3 + c
= (2/√3)*[atan(2x-1)] + c
= ∫dx/(x^2 - x + 1/4 + 3/4)
= ∫dx/[(x-1/2)^2 + 3/4]
= ∫d(x-1/2)/[(x-1/2)^2 + 3/4]
= ∫dy/(y^2 + 3/4) ;;; y = x - 1/2
= ∫dy/(y^2 + a^2) ;;; a = √3/2
= (1/a)∫d(y/a)/[(y/a)^2 + 1]
= (1/a)∫dz/(z^2 + 1) ;;; z = y/a
= [atan(z)]/a + c
= [atan(y/a)]*2/√3 + c
= [atan(2y/√3)]*2/√3 + c
= (2/√3)*[atan(2x-1)] + c
2015-10-12 10:04 pm
把題目看成(x^2-x+1)^-1
用-1次方積分公式
(分子微分*分母不微分+分母微分*分子不微分)/分母平方
得答案(-2X+1)/(X^2-x+1)^2
老師應該會說不用展開吧
用-1次方積分公式
(分子微分*分母不微分+分母微分*分子不微分)/分母平方
得答案(-2X+1)/(X^2-x+1)^2
老師應該會說不用展開吧
收錄日期: 2021-04-30 20:07:38
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151012125852AAZUBBQ