A. $ {1 \over 3}{x^3} +x + \arctan x + C $
B. $ {1 \over 3}{x^3} - x - \arctan x + C $
C. $ -{1 \over 3}{x^3} +x + \arctan x + C $
D. $ {1 \over 3}{x^3} - x + \arctan x + C $
求-|||-(1) (int )_(-1)^1f(x)dx;-|||-(2)f(x)dx;-|||-(3) (int )_(3)^-1g(x)dx;-|||-(
(int )_(-1)^2x|x|dx=(int )_(-1)^1x|x|x+(int )_(1)^2x|x|x|=0+(int )_(1)^2(x)^2dx
定积分(int )_(-4)^4((e)^(x^2)sin x+2)dx=( )(int )_(-4)^4((e)^(x^2)sin x+2)dx=( )定
int dfrac (x+1)({x)^2+4x+13}dx
不定积分int dfrac (dx)(4{x)^2-9}=-|||-__.int dfrac (dx)(4{x)^2-9}=-|||-__int dfrac (
(int )_(0)^dfrac (pi {4)}dfrac (x)(1+cos 2x)dx=( ) .(int )_(0)^dfrac (pi {4)}dfr
int (1-x)/(sqrt (9-4x^2))dx，$\int \frac {1-x}{\sqrt {9-4x^{2}}}dx$，
(d)/(dx) int_(1)^x x ln (x^2 + 1) , dx = $\frac{d}{dx} \int_{1}^{x} x \ln (x^2 +
int dfrac (5x-1)({x)^2-x-2}dx= (.A.int dfrac (5x-1)({x)^2-x-2}dx= (B.int dfrac (
定积分(int )_(0)^1sqrt (2x-{x)^2}dx=( )(int )_(0)^1sqrt (2x-{x)^2}dx=( )(int )_(0