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，则换元后定积分的下限与上限分别为( ) (int )_(0)^ln 2sqrt ({e)^x-1}dx (int )_(0)^ln 2sqrt ({e)^x-1}dx(int )_(0)^ln 2sqrt ({e)^x-1}dx(int )_(0)^ln 2sqrt ({e)^x-1}dx

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

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

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

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

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

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

求定积分(int )_(1)^sqrt (3)dfrac (1)({x)^2sqrt (1+{x)^2}}dx求定积分

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

.计算下列定积分:-|||-(10) (int )_(1)^sqrt (3)dfrac (dx)({x)^2sqrt (1+{x)^2}}

定积分(int )_(0)^1sqrt (2x-{x)^2}dx=( )(int )_(0)^1sqrt (2x-{x)^2}dx=( )(int )_(0