[题目]设 (x)=sqrt (1+{ln )^2x} 则 (e)=()-|||-A、 dfrac (sqrt {2)}(4)-|||-B、 dfrac (sqrt {2)}(2e)-|||-C、 dfrac (sqrt {2)}(2)e-|||-D、 sqrt (2)e

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

[题目]设函数 =ln dfrac (sqrt {{x)^2+1}}(sqrt [3]{x+2)}(xgt -2), 则 (0)= ()-|||-

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

(19) =ln ((e)^x+sqrt (1+{e)^2x}).

已知f(x)=(2x)/(sqrt (1-{x)^2)}，则(df(sqrt (1-{x)^2)})/(dx)=(,,,,,)A、-2；B、-dfrac(2x)

求lim _(xarrow infty )dfrac (ln (x+sqrt {{x)^2+1)}-ln (x+sqrt ({x)^2-1})}({({e)^d

(int )_(1)^(e^2)dfrac (ln x)(sqrt {x)}dx= __

[题目]-|||-int (dfrac (3)(1+{x)^2}-dfrac (2)(sqrt {1-{x)^2}})dx

.int dfrac (sqrt {1+{x)^2}+sqrt (1-{x)^2}}(sqrt {1-{x)^4}}dx.

4 单选 -|||-设函数 =ln dfrac (sqrt {{x)^2+1}}(sqrt [3]{x+2)}(xgt -2), 则 (0)=-|||-