下列凑微分的形式中错误的是( )A dfrac (1)(sqrt {{a)^2-(x)^2}}dx=d(arctan dfrac (x)(a))B dfrac (1)(sqrt {{a)^2-(x)^2}}dx=d(arctan dfrac (x)(a))C dfrac (1)(sqrt {{a)^2-(x)^2}}dx=d(arctan dfrac (x)(a))D dfrac (1)(sqrt {{a)^2-(x)^2}}dx=d(arctan dfrac (x)(a))

(14) int dfrac ({x)^2}(sqrt {{a)^2-(x)^2}}dx;

2x(1+x)-(x)^2-|||-7.设函数f(x)可导且 (x)=dfrac (2x)(sqrt {{a)^2-(x)^2}} 求 dfrac (d)(dx

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

[题目] =dfrac (1)(sqrt {2-{x)^2}}+arcsin (dfrac (1)(2)x-1) 的定义域是 __

求不定积分int dfrac (1)(2x)sqrt (ln x)dx=（）.int dfrac (1)(2x)sqrt (ln x)dx=int dfrac

求下列定积分:-|||-(int )_(dfrac {sqrt {2)}(2)}^1dfrac (sqrt {1-{x)^2}}({x)^2}dx

(6) () =dfrac (1)(sqrt {1-{x)^2}} int dfrac (1)(sqrt {1-{x)^2}}dx=() .

计算积分(int )_(dfrac {1)(2)}^dfrac (3{2)}dfrac (dx)(sqrt {|x-{x)^2|}}.(本题满分6分)计算积分.

计算下列定积分.-|||-(int )_(dfrac {sqrt {2)}(2)}^1dfrac (sqrt {1-{x)^2}}({x)^2}dx -

∫dfrac ({(1-x))^2}(sqrt {x)}dx=＿＿＿.∫＿＿＿.