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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

n(n+1) =(1/3) { n(n+1)(n+2) - (n-1)n(n+1) } 1x2+2x3+3x4+...99x100 = 1x2 + (1/3) { (2x3x4 - 1x2x3) + (3x4x5 - 2x3x4) +...+(99x100x101 - 98x99x100) } = 1x2 + (1/3) { 99x100x101 -1x2x3 } = (1/3) 99x100x101 =333300

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 .

3x-1/4=3/4 解:3x=3/4+1/4 3x=1 x=1/3

解: 原式X(X-1)=3X-3 x^2-x=3x-3(分解因式) x^2-4x+3=0(合并同类项) 解出:x=1或x=3

4/5

(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

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。

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

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...

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