x+1 (1+x 1+x2(1+ d(1+x d(1+x 2J(1+x d x In 1+x+ 2(1+x2)J(1+x2)2 x= tan t cOs、1 +cos 2t)dt (1+x t+-sin 2t+C=-arctan x +c 2(1+x2) 则 2x+x dx=In 1+x arctan]+C 2-1+x26 3 2 2 2 2 2 2 1 2 1 , [ ] (1 ) 1 (1 ) x x x x dx dx x x x         于是   2 2 2 2 2 2 2 (1 ) 1 (1 ) 1 2 (1 ) (1 ) d x d x dx x x x            2 2 2 2 1 ln 1 2(1 ) (1 ) dx x x x        2 2 2 1 tan cos (1 cos 2 ) (1 ) 2 dx x t tdt t dt x     而   2 1 1 1 sin 2 arctan 2 4 2 2(1 ) x t t C x C x        3 2 2 2 2 2 1 1 1 ln 1 [ arctan ] (1 ) 2 1 x x x dx x x C x x           则 