∫(1/(x^2+2x+5))dx的不定积分为1/2arctan((x+1)/2)+C
解:∫(1/(x^2+2x+5))dx
=∫1/[(x+1)^2+4]dx
=1/4∫1/[((x+1)/2)^2+1]dx
令(x+1)/2=t,则x=2t-1
则1/4∫1/[((x+1)/2)^2+1]dx
=1/4∫1/(t^2+1)d(2t+1)
=1/2∫1/(t^2+1)dt
=1/2arctant+C
把t=(x+1)/2代入,得
∫(1/(x^2+2x+5))dx=1/2arctan((x+1)/2)+C
扩展资料:
1、不定积分的公式类型
(1)含a+bx的不定积分
∫(1/(ax+b))=1/b*ln|ax+b|+C、∫(x/(ax+b))=1/b^2*(a+bx-aln|ax+b|)+C
(2)含x^2±a^2的不定积分
∫(1/(x^2+a^2))=1/a*arctan(x/a)+C、∫(1/(x^2-a^2))=1/(2a)*ln|(x-a)/(x+a)|+C
(3)含ax^2±b的不定积分
∫(1/(a*x^2+b))=1/√(a*b)*arctan(√a*x/√b)+C
2、不定积分的求解方法
(1)换元积分法
例:∫e^(2x)dx=1/2∫e^(2x)d(2x)=1/2*e^(2x)+C
(2)积分公式法
例:∫e^xdx=e^x、∫1/xdx=ln|x|+C、∫cosxdx=sinx+C
(3)分部积分法
例:∫x*e^xdx=∫xd(e^x)=x*e^x-∫e^xdx=x*e^x-e^x=(x-1)*e^x
参考资料来源:百度百科-积分公式
参考资料来源:百度百科-不定积分
积分公式∫(1\/(x^2+2x+5)) dx
∫(1\/(x^2+2x+5))dx的不定积分为1\/2arctan((x+1)\/2)+C 解:∫(1\/(x^2+2x+5))dx =∫1\/[(x+1)^2+4]dx =1\/4∫1\/[((x+1)\/2)^2+1]dx 令(x+1)\/2=t,则x=2t-1 则1\/4∫1\/[((x+1)\/2)^2+1]dx =1\/4∫1\/(t^2+1)d(2t+1)=1\/2∫1\/(t^2+1)...
x 1\/x^2-2x 5不定积分怎么求
∫(1\/(x^2+2x+5))dx的不定积分为1\/2arctan((x+1)\/2)+C 解:∫(1\/(x^2+2x+5))dx =∫1\/[(x+1)^2+4]dx =1\/4∫1\/[((x+1)\/2)^2+1]dx 令(x+1)\/2=t,则x=2t-1 则1\/4∫1\/[((x+1)\/2)^2+1]dx =1\/4∫1\/(t^2+1)d(2t+1)=1\/2∫1\/(t^2+1)...
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...
不定积分∫1\/(x^2+2x+5) dx的积分
那么,∫1\/(x^2+2x+5)dx =∫1\/((x+1)^2+4)dx =∫1\/((2tant)^2+4)d(2tant-1)=1\/4∫1\/(sect)^2d(2tant)=1\/2∫dt=t\/2+C 又因为x+1=2tant,所以t=arctan((x+1)\/2)则∫1\/(x^2+2x+5)dx=t\/2+C=1\/2*arctan((x+1)\/2)+C ...
∫1\/(x^2+2x+5)dx =∫1\/[(x+1)²+4] dx =∫1\/[(x+1)²+4] d(x+...
有一个积分公式 ∫ 1\/(x²+a²)dx =1\/a arctan x\/a 对应的x用x+1代替 a用2代替
不定积分(x+1)\/(x^2+2x+5)dx=
x^2+2x+5=(x+1)^2+4, 做个代换u=x+1,原式=∫u\/(u^2+4)du =1\/2∫ 1\/(u^2+4)d(u^2)=1\/2*ln(u^2+4)+C =1\/2*ln[(x+1)^2+4]+C =1\/2*ln(x^2+2x+5)+C
∫上限1下限-1[1\/(x平方+2x+5)]
∫1\/(x^2+2x+5) dx =∫1\/[(x+1)^2+4] dx =∫1\/4 *1\/[(x\/2+1\/2)^2+1] dx =1\/4* ∫ 1\/[(x\/2+1\/2)^2+1] d(x\/2+1\/2)=1\/4 *arctan(x\/2+1\/2) 代入上下限1和-1 =1\/4 *(arctan1 -arctan0)=1\/4 *π\/4 =π\/16 ...
积分号2x+2\/x^2+2x+5怎样换元法
∫[(2x+2)\/(x^2+2x+5)]dx =∫[1\/(x^2+2x+5)]d(x^2+2x+5)=ln(x^2+2x+5)2x+2提到dx 里面,其实就是d(x^2+2x)至于加多少对于积分没影响,即d(x^2+2x+C) C为任意常数。
换元法求不定积分1\/根号(x^2+2x+5)dx
原式=∫1\/√[(x+1)²+4]d(x+1)设x+1=2tant,t=actan[(x+1)\/2],则√[(x+1)²+4]=√[4(tan²t+1)]=√(4sec²t)=2sect,d(x+1)=2sec²tdt ∴原式=∫1\/√[(x+1)²+4]d(x+1)=∫1\/(2sect)*2sec²tdt =∫sectdt =ln|...