• 2022-06-15
    已知函数$y= \ln (1+ x) $,则$y''(x) =$( )。
    A: $\frac{1}{(1+x)^2}$
    B: $-\frac{1}{(1+x)^2}$
    C: $-\frac{1}{1+x}$
    D: $\frac{1}{1+x}$
  • B

    内容

    • 0

      下面积分收敛的是 A: $\int_0^\infty \frac{x^{4/3}}{1+x^2} dx$ B: $\int_1^\infty \frac{dx}{x \sqrt[3]{1+x^3}}$ C: $\int_1^\infty \frac{1}{x} dx$ D: $\int_1^\infty \frac{\arctan x}{x} dx$

    • 1

      方程${{x}^{2}}{{y}^{''}}-(x+2)(x{{y}^{'}}-y)={{x}^{4}}$的通解是( ) A: $y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{2}})$ B: $y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{4}})$ C: $y={{C}_{1}}x+{{C}_{2}}x{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{4}})$ D: $y={{C}_{1}}x+{{C}_{2}}x{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{2}})$

    • 2

      \(\int { { {\sec }^{3}}xdx}\)=( ) A: \(\frac{1}{2}\sec x\cot x-\frac{1}{2}\ln \left| \sec x+\tan x \right|+C\) B: \(\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln \left| \sec x+\tan x \right|+C\) C: \(-\frac{1}{2}\csc x\tan x+\frac{1}{2}\ln \left| \sec x-\cot x \right|+C\) D: \(-\frac{1}{2}\sec x\tan x-\frac{1}{2}\ln \left| \csc x+\tan x \right|+C\)

    • 3

      函数$f(x,y)={{\text{e}}^{-x}}\cos y$在点$(0,0)$处2次Taylor多项式为 A: $1+x+\frac{1}{2}({{x}^{2}}-{{y}^{2}})$ B: $1-x+\frac{1}{2}({{x}^{2}}-{{y}^{2}})$ C: $1-x+\frac{1}{2}({{x}^{2}}+{{y}^{2}})$ D: $1+x+\frac{1}{2}({{x}^{2}}+{{y}^{2}})$

    • 4

      1.下列函数中,在定义域上无界的函数是 A: $f(x)=\frac{1}{x}\sin x$ B: $f(x)=x^2\sin \frac{1}{x}$ C: $f(x)=\frac{\ln x}{1+{{\ln }^{2}}x}$ D: $f(x)=\frac{1}{{{\text{e}}^{x}}+{{\text{e}}^{-x}}}$