(int )_(0)^1dfrac (1)(2)xdx'= A.1B.0C.(int )_(0)^1dfrac (1)(2)xdx'=D.2

【题目】 (int )_(0)^dfrac (pi {2)}cos xdx= ()-|||-A、 dfrac (1)(2)-|||-B、1-|||-C、 -df

(int )_(0)^1dfrac (x+2)(sqrt {x+1)}dx=( )(int )_(0)^1dfrac (x+2)(sqrt {x+1)}dx=

(int )_(0)^1dfrac (dx)({x)^2sqrt (1-x)} 发散-|||-D (int )_(0)^1dfrac (dx)(xsqrt {1

(int )_(-1)^1dfrac (xdx)(sqrt {5-4x)}

(int )_(-1)^1dfrac (xdx)(sqrt {5-4x)}

证明 :(int )_(x)^1dfrac (dt)(1+{t)^2}=(int )_(1)^dfrac (1{x)}dfrac (dt)(1+{t)^2}(x

(11) (int )_(-1)^1dfrac (xdx)(sqrt {5-4x)};

(B)| (x/(1+2)^2 (1+x^2)^2 dx=0.-|||-(C) (int )_(-1)^1dfrac (1)(sin x)dx=0. (D) (

(int )_(0)^1dfrac (x)(sqrt {1-{x)^2}}dx= [填空1]

3.证明: (int )_(x)^1dfrac (dt)(1+{t)^2}=(int )_(1)^dfrac (1{x)}dfrac (dt)(1+{t)^2}