作者LuisSantos (^______^)
看板trans_math
标题Re: [积分]
时间Thu Feb 15 00:22:53 2007
※ 引述《king911015 (早已放弃爱上你)》之铭言：
: 1.
: x + 4
: ∫---------------- dx =?
: x^3 + x^2 - 2
x^3 + x^2 - 2 = (x - 1)(x^2 + 2x + 2)
x + 4 A Bx + C
令 --------------- = ------- + ---------------
x^3 + x^2 - 2 x - 1 x^2 + 2x + 2
x + 4 = (A)(x^2 + 2x + 2) + (x - 1)(Bx + C)
= (A + B)(x^2) + (2A - B + C)(x) + (2A - C)
A + B = 0 ------(1)
2A - B + C = 1 ------(2)
2A - C = 4 ------(3)
(1) => B = -A ------(4) 代入(2)
2A - (-A) + C = 1 => 3A + C = 1 ------(5)
(3) + (5) => 5A = 5 => A = 1 代入(3)、(4)
2 - C = 4 => C = 2 - 4 = -2 , B = -A = -1
x + 4 1 -x - 2
--------------- = ----- + ------------
x^3 + x^2 - 2 x - 1 x^2 + 2x + 2
x + 4
∫--------------- dx
x^3 + x^2 - 2
1 -x - 2
= ∫------- dx + ∫-------------- dx
x - 1 x^2 + 2x + 2
1 -1 2x + 4
= ∫------- dx + (---)(∫-------------- dx)
x - 1 2 x^2 + 2x + 2
1 -1 2x + 2 2
= ∫------- dx + (---)(∫-------------- dx) + ∫-------------- dx
x - 1 2 x^2 + 2x + 2 x^2 + 2x + 2
1 -1 2x + 2 1
= ∫------- dx + (---)(∫-------------- dx) + (2)(∫----------------- dx)
x - 1 2 x^2 + 2x + 2 (x + 1)^2 + 1^2
1 -1
= ln|x - 1| - (---)(ln(x^2 + 2x + 2)) + (2)(tan (x + 1)) + c
2
: 2.
: ∫ cos√x dx = ?
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