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3x 3 1 x 4

∫(3x^3+x)/(1+x^4)dx =3∫x^3/(1+x^4)dx+∫x/(1+x^4)dx =3/4∫1/(1+x^4)d(1+x^4)+1/2∫1/(1+x^4)dx^2 =2/3ln(1+x^4)+1/2arctanx^2+C

由 x+1<3x-3 1 2 (x-4)< 1 3 (x-4) 求得 2<x x<4 ,则2<x<4.解方程x 2 -2x-4=0可得x 1 =1+ 5 ,x 2 =1- 5 ,∵2< 5 <3,∴3<1+ 5 <4,符合题意∴x=1+ 5 .

解:4(X-1)+3≥3X 4x-4+3≥3x 4x-1-3x≥0 x-1≥0 x≥1

3x-1/4=3/4 解:3x=3/4+1/4 3x=1 x=1/3

因为n(n+1)=n平方+n 原式=(1平方+2平方+……n平方)+(1+2+3……n) =n(n+1)(2n+1)/6+n(n+1)/2为标准答案 注:Sn(1平方+2平方+……n平方)的证明: (a+1)³-a³=3a²+3a+1(即(a+1)³=a³+3a²+3a+1) a=1时:2³...

过程如下: ∫3x^4+3x^2+1/x^2+1dx =∫[3(x^4+x^2)+1]/x^2+1dx =∫3x^2dx+∫dx/x^2+1 =x^3+arctanx+C。

一般的,有: (n-1)n(n+1) =n^3-n {n^3}求和公式:Sn=[n(n+1)/2]^2 {n}求和公式:Sn=n(n+1)/2 1x2x3+2x3x4+3x4x5+....+7x8x9 =2^3-2+3^3-3+...+8^3-8 =(2^3+3^3+...+8^3)-(2+3+...+8) =[(8*9/2)^2-1]-8*9/2+1 =1260

1/2+1/2x3+1/3x4+…+1/2016x2017 =1-1/2+1/2-1/3+1/3-1/4+……+1/2016-1/2017 =1-1/2017 =2016/2017

(3x-4)/[(x+1)(x-2)]=A/x-1 + B/x-2. 则 A/x+1 + B/x-2=[A(x-2)+B(x+1)]/[(x+1)(x-2)] (A+B)x-2A+B=3x-4 A+B=3 (1) -2A+B=-4 (2) (1)*2+(2)得 3B=2 B=2/3 A=7/3

1x2+2x3+3x4+…+n(n+1) =1x(1+1)+2x(2+1)+3x(3+1)+…n(n+1) =(1^2+2^2+3^2+…+n^2)+(1+2+3+…+n) =n(n+1)(2n+1)/6+n(n+1)/2 =n(n+1)[(2n+1)+3]/6 1x2+2x3+3x4+…+10x11 =10x(10+1)x[(10x2+1)+3]/6 =110x4 =440

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