求不定积分∫1\/x^2+2x+3dx
∫ 1\/(x² + 2x + 3) dx = ∫ 1\/[(x + 1)² + 2] dx = (1\/√2)arctan[(x + 1)\/√2] + C 其中一条积分表公式:∫ dx\/(a² + x²) = (1\/a)arctan(x\/a) + C
1\/x^2+2x+3的不定积分?
回答:分母可以写成(x+1)^2 +2 然后套不定积分公式 1\/(x^2+a^2)
x\/(x的平方+2x+3)dx的不定积分是多少?多谢,详细步骤
∫x\/(x的平方+2x+3)dx =∫x\/[(x+1)²+2] dx =∫[(x+1)-1]\/[(x+1)²+2] dx =∫[(x+1)\/[(x+1)²+2] dx-∫1\/[(x+1)²+2] dx =1\/2∫1\/(x的平方+2x+3) d(x的平方+2x+3) -1\/√2 arctan(x+1)\/√2+c =1\/2ln(x的平方+2x+...
1\/(x^2+2x+3)的不定积分。谢谢啦
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求不定积分1\/x^2+2x+3
=½∫(2x-2)\/(x²+2x+3)dx =½∫(2x+2-4)\/(x²+2x+3)dx =½∫(2x+2)\/(x²+2x+3)dx - ½∫4\/(x²+2x+3)dx =½∫(2x+2)\/(x²+2x+3)dx - 2∫1\/(x²+2x+3)dx =½∫d(x²+2x+3)\/(x&...
求1\/(x^2+2x+3)的不定积分.详细过程,谢谢
x^2+2x+3=(x+1)^2+2,原积分=∫dx\/(x+1)^2+2 =(1\/√2)∫d[(x+1)\/√2]\/{[(x+1)\/√2]^2+1} =(1\/√2)arctg[(x+1)\/√2]+C
∫(x+1)\/(x^2+2x+3)dx不定积分怎么求啊各位大神
∫(x+1)\/(x^2+2x+3)dx=1\/2∫2(x+1)\/(x^2+2x+3)dx =1\/2∫1\/(x^2+2x+3)d(x^2+2x+3)=1\/2ln(x^2+2x+3)
用第一换元法求不定积分1\/x^2+2x-3dx
2017-12-08 求∫x-1\/x∧2 2x 3dx 3 2017-03-15 ∫x^2\/(x+2)^3dx 用第一换元法求解,详细过程 7 2010-11-29 用换元法求不定积分 ∫1\/(2x-3)^2dx 2 2015-06-28 不定积分1\/根号(1-x^2)^3dx 用第二换元积分法求 3 2016-04-24 用一类换元法求∫x(1+2x^2)^05dx不定积分 ...
求不定积分∫(1\/x^2+2x+5)dx
结果为:(1\/2)arctan[(x+1)\/2]+ C 解题过程如下:原式=∫1\/(x^2+2x+5)dx =∫1\/[(x+1)^2+4]dx =∫(1\/4)\/[ [(x+1)\/2]^2+1]dx =∫(1\/4)·2\/[ [(x+1)\/2]^2+1]d( (x+1)\/2)=(1\/2)∫1\/[ [(x+1)\/2]^2+1]d( (x+1)\/2)=(1\/2)arctan[(x+1...