Re: [微積] 問分式積分~
※ 引述《SoDaLiBou (SoDaStyle)》之銘言：
: 請問各位大大
: (2+2x)/(1-2x-x^2)
: 這式子該怎麼積分
: 我問朋友她說令分母為u
: 可是我還是不懂
: 想問有人可以詳細解一下嗎?
: 感激
2+2x 2+2x 2+2x
∫────dx = - ∫ ──── dx = -∫ ───────── dx
1-2x-x^2 x^2+2x-1 (x+1+√2)(x+1-√2)
1 1
= -∫──── dx -∫────dx
x+1+√2 x+1-√2
= -ln│x+1+√2│ - ln │x+1-√2│ +c
Explain:
2+2x A B
───────── = ───── + ──────
(x+1+√2)(x+1-√2) (x+1+√2) (x+1-√2)
use Heaviside Cover-up Method, than we could get A=1 and B=1
Other Method:
let u= x^2 +2x-1
du = (2x+2)dx
2x+2 1
- ∫──── dx = - ∫ ─── du = - ln│u│+c = -ln│x^2+2x-1│ +c
x^2+2x-1 u
--
Logic can be patient for it is eternal. ----- Oliver Heaviside
--
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※ 文章網址: http://www.ptt.cc/bbs/Math/M.1415633364.A.DC1.html
※ 編輯: Heaviside (111.185.133.106), 11/10/2014 23:33:08
推
11/10 23:40, , 1F
11/10 23:40, 1F
→
11/11 07:28, , 2F
11/11 07:28, 2F
→
11/11 07:29, , 3F
11/11 07:29, 3F
推
11/11 14:10, , 4F
11/11 14:10, 4F
感謝更正 每次從計算紙寫上來都容易抄錯= ="
※ 編輯: Heaviside (111.185.133.106), 11/11/2014 17:13:32
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