Re: [積分]

看板trans_math作者 (^______^)時間19年前 (2007/03/27 20:00), 編輯推噓1(100)
留言1則, 1人參與, 最新討論串128/259 (看更多)
※ 引述《gwish (曖曖內含光)》之銘言: : 2x+2 : ∫--------------dx : (x-1)^3(x^2+1) : 這題部分分式一直算錯 : 不知道是不是哪裡搞錯了 : 謝謝 2x + 2 A B C Dx + E 令 -------------------- = ----- + ------- + ------- + -------- ((x-1)^3)(x^2 + 1) x - 1 (x-1)^2 (x-1)^3 x^2 + 1 則 2x + 2 = (A)((x-1)^2)(x^2 + 1) + (B)(x-1)(x^2 + 1) + (C)(x^2 + 1) + (Dx + E)((x-1)^3) = (A)(x^2 - 2x + 1)(x^2 + 1) + (B)(x-1)(x^2 + 1) + (C)(x^2 + 1) + (Dx + E)(x^3 - (3)(x^2) + 3x - 1) = (A)(x^4 - (2)(x^3) + (2)(x^2) - 2x + 1) + (B)(x^3 - x^2 + x - 1) + (C)(x^2 + 1) + (D)(x^4) + (-3D+E)(x^3) + (3D-3E)(x^2) + (-D+3E)(x) - E = (A + D)(x^4) + (-2A + B - 3D + E)(x^3) + (2A - B + C + 3D - 3E)(x^2) + (-2A + B - D + 3E)(x) + (A - B + C - E) = 2x - 2 A + D = 0 ------(1) -2A + B - 3D + E = 0 ------(2) 2A - B + C + 3D - 3E = 0 ------(3) -2A + B - D + 3E = 2 ------(4) A - B + C - E = 2  ------(5) (1) => D = -A ------(6) (6)代入(2)、(3)、(4)得 A + B + E = 0 ------(7) -A - B + C - 3E = 0 ------(8) -A + B + 3E = 2 ------(9) (7) => E = -A - B ------(10) (10) 代入 (5)、(8)、(9)得 A - B + C + A + B = 2 => 2A + C = 2 ------(11) -A - B + C + 3A + 3B = 0 => 2A + 2B + C = 0 ------(12) -A + B - 3A - 3B = 2 => 4A + 2B = -2 => 2A + B = -1 ------(13) (11) => C = 2 - 2A 代入(12) 2A + 2B + 2 - 2A = 0 => 2B + 2 = 0 => B + 1 = 0 => B = -1 代入 (13) 2A - 1 = -1 => 2A = 0 => A = 0 代入(11)得 C = 2 A = 0 代入(6)得 D = -A = 0 A = 0 , B = -1 代入(10)得 E = -A - B = 0 - (-1) = 0 + 1 = 1 2x + 2 -1 2 1 ------------------ = ------- + ------- + ------- ((x-1)^3)(x^2 + 1) (x-1)^2 (x-1)^3 x^2 + 1 2x + 2 ∫------------------ dx ((x-1)^3)(x^2 + 1) -1 2 1 = ∫------- + ------- + ------- dx (x-1)^2 (x-1)^3 x^2 + 1 1 -1 1 -1 = (-1)(-1)(-----) + (2)(---)(-------) + tan x + c x - 1 2 (x-1)^2 1 1 -1 = ----- - ------- + tan x + c x - 1 (x-1)^2 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.119.27.61

03/28 00:08, , 1F
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03/28 00:08, 1F
文章代碼(AID): #162GTnev (trans_math)
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