曲线=ln (1-(x)^2)在=ln (1-(x)^2)上的一段弧长为（ ）。=ln (1-(x)^2)=ln (1-(x)^2)=ln (1-(x)^2)=ln (1-(x)^2)

设=ln sqrt (dfrac {1-x)(1-{x)^2}}则 dy|=ln sqrt (dfrac {1-x)(1-{x)^2}}设则dy|

函数(x)=dfrac (ln |x|)(sqrt {1-{x)^2}}的定义域是 A(x)=dfrac (ln |x|)(sqrt {1-{x)^2}} B

=dfrac (arcsin x)(x)+dfrac (1)(2)ln dfrac (1-sqrt {1-{x)^2}}(1+sqrt {1-{x)^2}}

(int )_(1)^edfrac (dx)(xsqrt {1-{(ln x))^2}}..

极限 lim _(xarrow 0)dfrac ({(2+cos x))^x-(3)^x}((1-sqrt {1-{x)^2})ln (1+} 为() ()

求下列函数的二阶导数:-|||-(1) =(cos )^2xcdot ln x;-|||-(2) =dfrac (x)(sqrt {1-{x)^2}}

补考-|||- +ln |x+2|+C 则常数 int dfrac (1)(5x)dx= ()()-|||-A ln |5x|+C-|||-B 1-|||-C

曲线=dfrac (x)(1-{x)^2}的渐近线是（ ）.=dfrac (x)(1-{x)^2}和=dfrac (x)(1-{x)^2}=dfr

求极限 lim _(x arrow 0) (x- ln (1+x))/(1- cos x)17. 求极限 $\lim _{x \rightarrow 0} \f

=(ln )^2(1-x),则=(ln )^2(1-x)________.,则________.