5∫5 x³sin2x x4 2x2 1d dx

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∫(x+1)/(x^2-2x+5)dx

∫(x+1)/(x^2-2x+5)dx=1/2*∫(2x-2)/(x^2-2x+5)dx+∫2/(x^2-2x+5)dx=1/2*∫[1/(x^2-2x+5)]d(x^2-2x+5)+2∫1/[(x-

x^5+x^4 = (x^3-x)(x^2+x+1)+x^2+x

是这样的:x^5+x^4=x^3(x^2+x)=(x^2+x)[(x^3-1)+1]=(x^2+x)(x^3-1)+x^2+x=[x(x+1)(x-1)](x^2+x+1)+x^2+x=(x^3-x)

∫x√(x-5)dx

令t=√(x-5)去求解x=5+t^2dx=2tdt原积分=∫(5+t^2)*t*2tdt=∫(10t^2+2t^4)dt=10/3*t^3+2/5*t^5+c将t=√(x-5)代回结果即可得到结果.

|X-1|+|X-2|+|X-3|+|X-4|+|X-5|+|X-6|+|X-7|+|X-8|+|X-9|+|X-10|

|x-1|+|x-10|表示数轴上x到1的距离+x到10的距离.显然最小值是9,此时x只要在1到10之间就好.类似的,|x-2|+|x-9|的最小值是7,此时x在2到9之间就好.|x-3|+|x-8|

x+2x+3x+4x+5x+6x+7x+8x+9x=9x-8x-7x-6x-5x-4x-3x-2x-x.x等于多少?

x(1+2+...+9)=x(9-8-7-...-1)x=0记得采纳啊

∫2x²+3x-5/x+3dx

设x+3=t→dx=dt,代入原式得∫[(2x²+3x-5)/(x+3)]dx=∫[(2(t-3)²+3(t-3)-5)/t]dt=∫[2t+(4/t)-9]dt=t²+

∫(x^5+x^4-8)/x^3-x dx求不定积分具体过程

∫(x^5+x^4-8)/(x^3-x)dx=S(x^5-x^3+x^4-x^2+x^3-x+x^2-1+x-1-6)(x^3-x)dx=S(x^2+x+1+1/x+1/x(x+1)-6/(x^3-x

∫[(x+1)/(x^(2)+2x+5]dx

原式=0.5∫d(x²+2x+5)/(x²+2x+5)=0.5㏑(x²+2x+5)

∫(5x^2-x+1)/(x^3-4x^2)dx

这种形式的积分有一种通用的解法:待定系数法,看图你就明白了!再答:

∫(x^4-4x^2+5x-15)/(x^2+1)(x-2) dx=?

∵(x^4-4x^2+5x-15)/[(x^2+1)(x-2)]=[(x^4+x²-5x²-5)+(5x-10)]/[(x²+1)(x-2)]=[x²(x&su

x/x-3-5x/x^2-x-6计算

原式=[x/(x-3)]-[5x/(x²-x-6)]=[x/(x-3)]-(5x)/[(x-3)(x+2)]=[x(x+2)]/[(x-3)(x+2)]-(5x)/[(x-3)(x+2)]=

∫4x-3√x-5/x*dx求解

每一个分出来积分,答案是2x^2-2x^(3/2)-5lnx

∫x/(x^2+5)dx

∫x/(x^2+5)dx=1/2(ln|x^2+5|)+C

不定积分∫(3^x*5^x/25^x-9^x)dx

∫3^x×5^x/(25^x-9^x)dx=∫3^x×5^x/[(5²)^x-(3²)^x]dx=∫[1/(5^x-3^x)-1/(5^x+3^x)]×5^xdx=∫{1/[(5/

∫ [(x^3-2x^2+x+1)/(x^4+5x^2+4)]dx

[(x^3-2x^2+x+1)/(x^4+5x^2+4)]=1/(x^2+1)+(x-3)/(x^2+4).原式=∫1/(x^2+1)dx+∫(x-3)/(x^2+4)dx=arctanx+(1/2)

用递归法求y=x-(x*x*x/3!)+(x*x*x*x*x/5!)-(x*x*x*x*x*x*x/7!)+.

#includesintjiesheng(intn)//用于计算阶乘如:3!{if(n=1)rerunn;returnn*hh(n-1)}doubledigui(intn,intx)//用以计算y=x

∫dx/x(x^5+4)

∫1/[x(x^5+4)]dx=¼∫[(x^5+4)-x^5]/[x(x^5+4)]dx=¼∫[1/x-x^4/(x^5+4)]dx=¼[∫1/xdx-1/5∫1/(x^

x+2/x+1-x+3/x+2-x+4/x+3+x+5/x+4

/>(x+2)/(x+1)-(x+3)/(x+2)-(x+4)/(x+3)+(x+5)/(x+4)=1+1/(x+1)-1-1/(x+2)-1-1/(x+3)+1+1/(x+4)=1/(x+1)-1/

通分x+4/x*x+4x与x-5/x(x-5)-2(x-5)

不太明白你的意思,是二个多项式通分同分母吗?(x+4)/(x^2+4x)=(x+4)/[x(x+4)]=1/x=(x-2)/[x(x-2)(x-5)/[x(x-5)-2(x-5)]=(x-5)/[(x