解:」 dx 0 dx + dx 1 2 2 十x +x2101+x2 0 b m i dx tlim a→-0 1+x 0 2 b→》+ lim arctan xla lim arctan x Jo a→-0 b→)+∞ =-lim arcta na+ lim arctan T T T 2)2 上页 + − + 2 1 x d x − + = 0 2 1 x d x  + + + 0 2 1 x d x  + = →−  0 2 1 1 lim a a d x x  + + →+  b b d x x 0 2 1 1 lim   0 lim arctan a a x →−  =   b b arctan x 0 lim →+  + a a lim arctan →−  = − b b lim arctan →+  + . 2 2 =    +       = − − 解: