(B) ^(x^4-2x)-1.-|||-(C) (int )_(0)^(x^2)sin (t)^2dt. (D) sqrt (1+2x)-sqrt [3](1+3x)

计算下列各导数：(1) (d)/(dx)int_(0)^x^2sqrt(1+t^2)dt;(2) (d)/(dx)int_(x^2)^x^3(dt)/(sqrt

int_(1)^sqrt(3) dfrac(d x)(x^2 sqrt(1+x^2))$ \int_{1}^{\sqrt{3}} \dfrac{d x}{x^{

设xarrow 0时, f(x)=sqrt(1+2x)-sqrt[3](1+3x)的等价无穷小是ax^b, 则(a,b)=【】A. $\left(\frac{1

设f(x)连续，xgt 0，且(int )_(1)^(x^2)f(t)dt=(x)^2(1+x)，则f(2)=(,,,,,)A、4B、2sqrt (2)+12C

当 x arrow 0 时，下列无穷小量中阶数最高的是 ( )(A) int_(x^2)^x^3 sqrt(1 - sqrt(cos t)) , dt(B) i

lim _(xarrow 4)dfrac (sqrt {1+2x)-3}(x-4)= (-|||-A dfrac (2)(3) .-|||-B 2-|||-C

lim _(xarrow 0)dfrac (sin 2x-2sin x)(x(sqrt [3]{1+{x)^2}-1)}..

(4) lim _(xarrow 0)dfrac (sin x-tan x)((sqrt [3]{1+{x)^2}-1)(sqrt (1+sin x)-1)}

分) 设 (x)=(int )_({x)^2}dfrac (t)(sqrt {1+{t)^3}}dt 求 =(int )_(0)^1xF(x)dx

定积分(int )_(0)^1sqrt (2x-{x)^2}dx=( )(int )_(0)^1sqrt (2x-{x)^2}dx=( )(int )_(0