求下列反常积分的值：
(1)∫0+∞xe-x2dx；
(2)∫-∞+∞[1/(x2+2x+2)]dx；
(3)∫1+∞dx/x(1+x2)；
(4)∫+∞2/π(1/x2)sin(1/x)dx．
(1)∫0+∞xe-x2dx =1/2∫0+∞e-x2dx2 =-(1/2)e-x2|0+∞ =1/2 (2)∫-∞+∞[1/(x2+2x+2)]dx =∫-∞+∞1/[1+(x+1)]2(x+1) =arctan(x+1)|-∞+∞ =π (3)∫1+∞dx/x(1+x2) =∫1+∞(1/x-x/(1+x)2)dx =lnx√(x2+1)|1+∞ =(1/2)ln2． (4)∫2/π+∞(1/x2)sin(1/x)dx =∫2/π+∞sin(1/x)d(1/x) =cos(1/x)|2/π+∞ =1
