设=ln (x)^2+ln 2x，则=ln (x)^2+ln 2x（ ）=ln (x)^2+ln 2x=ln (x)^2+ln 2x=ln (x)^2+ln 2x=ln (x)^2+ln 2x

=ln (1+(e)^2x)+ln 2

设函数f(x)=x^2+ln(2-x)，则f(1)=1。( )设函数$$f(x)=x^2+ln(2-x)$$，则$$f(1)=1$$。( )

lim _(xarrow 0)dfrac (ln ({sin )^2x+(e)^x)-x}(ln ({x)^2+(e)^2x)-2x}

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

计算定积分(int )_(1)^edfrac (2+ln x)(x)dx.计算定积分.

int dfrac (1)(x{ln )^2x}dx

计算给出函数的n导数:-|||-(1) =(e)^xcdot (x)^2-|||-(2) =(x)^2+ln (1+x)+(2)^x

6.求下列极限:-|||-lim _(xarrow 0)dfrac (ln ({sin )^2x+(e)^x)-x}(ln ({e)^2x-(x)^2)-2x}

(int )_(e)^+infty dfrac (dx)(x{ln )^2x}= __

(2) lim _(xarrow 0)[ dfrac (1)(ln (1+{tan )^2x)}-dfrac (1)(ln (1+{x)^2)}]