1.判断题:-|||-(1)因为sinx为奇函数,所以 (int )_(-infty )^+infty sin xdx=0. ()-|||-(2) (int )_(-infty )^+infty dfrac (2x)(1+{x)^2}dx=lim _(aarrow +infty )(int )_(-a)^adfrac (2x)(1+{x)^2}dx=0. ()-|||-(3) (int )_(0)^4dfrac (dx)({(x-3))^2}=-dfrac (1)(x-3)(|)^4=-dfrac (4)(3) ()