作者LuisSantos (^______^)
看板trans_math
標題Re: [積分]請問一些部份積分
時間Fri Oct 5 17:42:46 2007
※ 引述《kissmeeggy (eggy)》之銘言：
: 7-x-2x^2
: 一: ∫---------------- dx
: (x-1)^2(x^2+x+2 )
7 - x - (2)(x^2) A B Cx + D
令 ---------------------- = ----- + --------- + -----------
((x-1)^2)(x^2 + x + 2) x - 1 (x - 1)^2 x^2 + x + 2
7 - x - (2)(x^2) = (A)(x-1)(x^2 + x + 2) + (B)(x^2 + x + 2)
+ (Cx + D)((x - 1)^2)
= (A + C)(x^3) + (B - 2C + D)(x^2) + (A + B + C - 2D)(x)
+ (-2A + 2B + D)
A + C = 0 => C = -A ------(1)
B - 2C + D = -2 ------(2)
A + B + C - 2D = -1 ------(3)
-2A + 2B + D = 7 ------(4)
(1)代入(2)、(3)得
2A + B + D = -2 ------(5)
B - 2D = -1 => B = 2D - 1 ------(6)
(6)代入(4)、(5)得
-2A + 4D - 2 + D = 7 => -2A + 5D = 9 ------(7)
2A + 2D - 1 + D = -2 => 2A + 3D = -1 ------(8)
(7) + (8) => 8D = 8 => D = 1 代入(7)、(6)得
-2A + 5 = 9 => 2A = -4 => A = -2 => C = -A = 2
B = 2D - 1 = 2 - 1 = 1
-7 - x - (2)(x^2) -2 1 2x + 1
------------------------ = ----- + --------- + -----------
((x - 1)^2)(x^2 + x + 2) x - 1 (x - 1)^2 x^2 + x + 1
-7 - x - (2)(x^2)
∫------------------------ dx
((x - 1)^2)(x^2 + x + 2)
-2 1 2x + 1
= ∫----- + --------- + ----------- dx
x - 1 (x - 1)^2 x^2 + x + 1
1
= (-2)(ln|x - 1|) - ----- + ln|x^2 + x + 1| + c
x - 1
1
= (-2)(ln|x - 1|) - ----- + ln(x^2 + x + 1) + c
x - 1
1 3 3
(x^2 + x + 1 = (x + ---)^2 + --- ≧ ---)
2 4 4
: x^3+4x^2-4x-1
: 二: ∫-----------------dx
: (x^2+1)^2
: x
: 三: ∫---------------dx
: (x+1)^2(x^2+1)
: x^2+2x-1
: 四: ∫-------------dx
: 2x^3+3x^2-2x
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