[2018年] 曲线y=x2+2lnx在其拐点处的切线方程是________.

求下列不定积分:-|||-19. int (e)^sqrt [3](x)dx..

求指导本题解题过程，谢谢您！设有线性方程组 ) a(x)_(1)+2(x)_(2)+3(x)_(3)=8 2a(x)_(1)+2(x)_(2)+3(x)_(3)=10 (x)_(1)+(x)_(2)+b(x)_(3)=5 .-|||-(1)讨论a,b为何值时,方程组有唯一解,无-|||-解,有无穷多解;-|||-(2)在方程组有无穷多解时用方程组对应的-|||-齐次线性方程组的基础解系表示其通解

.A=A= (& -dfrac (1)(12) 0& 4& -24 0& 0& dfrac (1)(5) -12 6 0-|||--2 -5 4

3.求由 (int )_(0)^y(e)^tdt+(int )_(0)^xcos tdt=0 所确定的隐函数对x的导数 dfrac (dy)(dx) .

设G=<a>是15阶循环群,求(1) G的所有生成元; (2) G的所有子群。

设随机变量X的分布律为p dfrac (-1)({(a-1))^2(1-a)(1+a) 1-a}求：（1）a的值；（2）X的分布函数F(x).

1.设随机向量(X,Y)的分布函数在 leqslant xleqslant dfrac (pi )(2) leqslant yleqslant dfrac (pi )(2) 时为 (x,y)=sin xsin y _-|||-求数学期望E(X ),E(Y)及方差D(X ),D(Y),协方差Cov(X,Y).

第四章 不正积分-|||-(20)arctan√x/dx;-|||-(10) sqrt (18sqrt {1+{x)^2}}cdot dfrac (xsqrt {x)}(sqrt {1+{x)^2}}-|||-(22) int dfrac (dx)(sin xcos x)-|||-(21) int dfrac (1+ln x)({(xln x))^2}dx-|||-(24)∫cos^3xdx-|||-(23) int dfrac (ln sin x dx)(cos xsin x)dx-|||-(26)∫sin2xcos33xdx;-|||-(25) int (cos )^2(omega t+varphi )dt;-|||-(28)int sin 5xsin 7xdx;-|||-(27) int cos xcos dfrac (x)(2)dx;-|||-(29) int (sin )^3xsec xdx;-|||-(30) int dfrac (dx)({e)^x+(e)^-x}-|||-(31) int dfrac (1-x)(sqrt {9-4{x)^2}}dx-|||-(32) int dfrac ({x)^3}(9+{x)^2}dx;-|||-(33) int dfrac (dx)(2{x)^2-1};-|||-(34) int dfrac (dx)((x+1)(x-2));-|||-(35) int dfrac (x)({x)^2-x-2}dx;-|||-(36) int dfrac ({x)^2dx}(sqrt {{a)^2-(x)^2}}(agt 0)-|||-(37) int dfrac (dx)(xsqrt {{x)^2-1}}-|||-(38) int dfrac (dx)(sqrt {{({x)^2+1)}^3}}-|||-(39) int dfrac (sqrt {{x)^2-9}}(x)dx-|||-(40) int dfrac (dx)(1+sqrt {2x)}-|||-(41) int dfrac (dx{d)^2}(1+sqrt {1-{x)^2}}-|||-(42) int dfrac (dx)(x+sqrt {1-{x)^2}}-|||-(43) int dfrac (x-1)({x)^2+2x+3}dx;-|||-(44) int dfrac ({x)^3+1}({({x)^2+1)}^2}dx.-|||-前面我们在复合函数求导法则的基础上,得到了换元积分法.-|||-个函数乘积的求导法则,来推得另一个求积分的基本方法一-|||-设函数 u=u(x) 及 v=v(x) 具有连续导数,则两个函数乘积的-|||-(uv)'=u'v+uv'

判断向量组1 2-|||-x1= 2 ,α2= -1-|||-2 11 2-|||-x1= 2 ,α2= -1-|||-2 1的线性相关性。（）A.线性无关B.线性相关C.无法确定D.可相关也可无关

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