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

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

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 (x)(sqrt {1-{x)^2}}，则=dfrac (x)(sqrt {1-{x)^2}}=_________.，则=_________.

设=dfrac (arcsin x)(sqrt {1-{x)^2}}（1）证明：=dfrac (arcsin x)(sqrt {1-{x)^2}}（2）求=df

=dfrac (sqrt {1+x)-sqrt (1-x)}(sqrt {1+x)+sqrt (1-x)}

(9) =dfrac (sqrt {1+x)-sqrt (1-x)}(sqrt {1+x)+sqrt (1-x)} ;

计算(int )_(0)^1dx(int )_(1-x)^sqrt (1-{x^2)}dfrac (x+y)({x)^2+(y)^2}dy=-|||-dv=__

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

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

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