(int )_(0)^2dfrac (dx)({(1-x))^2};;
(int )_(0)^2dfrac (dx)({(1-x))^2};
求(int )_(0)^2dfrac (dx)({(1-x))^2}求
(int )_(1)^2dfrac (dx)(x(1+sqrt {2x))}=( )A.(int )_(1)^2dfrac (dx)(x(1+sqrt {2x
(int )_(0)^1dfrac (dx)({x)^2sqrt (1-x)} 发散-|||-D (int )_(0)^1dfrac (dx)(xsqrt {1
(int )_(1)^2dfrac (sqrt {{x)^2-1}}(x)dx
int dfrac ({(1-x))^2}(x(1+{x)^2)}dx
int dfrac (1)({x)^2}sqrt (dfrac {1-x)(1+x)}dx
例: (int )_(0)^dfrac (pi {2)}(cos )^2dfrac (x)(2)dx= __ .
不定积分int dfrac (1)(1+sqrt {1-x)}dx=( ) int dfrac (1)(1+sqrt {1-x)}dx=int dfrac (