(5) (int )_(0)^ln 2sqrt ({e)^x-1}dx

计算(int )_(0)^ln 2sqrt ({e)^x-1}dx时，常用换元法，若令(int )_(0)^ln 2sqrt ({e)^x-1}dx，则换元后定

求int dfrac ({ln )^2sqrt (x)}(sqrt {x)}dx求

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

(8) (int )_(0)^1(x)^2sqrt (1-{x)^2}dx :

(1) (int )_(0)^a(x)^2sqrt ({a)^2-(x)^2}dx(agt 0);

设 (int )_(0)^xuf(u)du=(e)^x-1, 则 (int )_(0)^sqrt (ln 2)(x)^3f((x)^2)dx= ()-|||-(

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

(5)int ln^2(x+sqrt(1+x^2))dx;(5)$\int ln^{2}(x+\sqrt{1+x^{2}})dx;$

(int )_(1)^sqrt (3)dfrac (dx)({x)^2sqrt (1+{x)^2}};

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