=dfrac (arcsin x)(x)+dfrac (1)(2)ln dfrac (1-sqrt {1-{x)^2}}(1+sqrt {1-{x)^2}}
若 (x)=ln (x+sqrt (1+{x)^2}) 则 (x)=dfrac (1)(sqrt {1+{x)^2}}-|||-5-|||-○ C.若 (x)=
讨论下列函数的有界性:(1) (x)=dfrac (x)(1+{x)^2}-|||-(2) (x)=dfrac (1+{x)^2}(1+{x)^4}讨论下列函数
=dfrac (x)(sqrt {1-{x)^2}},则=dfrac (x)(sqrt {1-{x)^2}}=_________.,则=_________.
不定积分int dfrac (1)(1+sqrt {x)}dx= __ 。() int dfrac (1)(1+sqrt {x)}dx= __ 。
(int )_(1)^2dfrac (dx)(x(1+sqrt {2x))}=( )A.(int )_(1)^2dfrac (dx)(x(1+sqrt {2x
不定积分int dfrac (1)(1+sqrt {1-x)}dx=( ) int dfrac (1)(1+sqrt {1-x)}dx=int dfrac (
(19) int tan sqrt (1+{x)^2}cdot dfrac (xdx)(sqrt {1+{x)^2}}-|||-(20) int dfrac (
(19) int tan sqrt (1+{x)^2}cdot dfrac (xdx)(sqrt {1+{x)^2}}-|||-(20) int dfrac (
int tan sqrt (1+{x)^2}cdot dfrac (xdx)(sqrt {1+{x)^2}}