例6求「 arctan xdx 解」 arctan xdx= arctan xd 2 x arctan. 2了21+x2 2mx2J(1-1+x2 arctanx-(x-arctan x)+C x+1=x+Carctan .  求 x x dx  xarctan x dx  = 2 arctan 2 x x d  + =  − x x x x x d 2 2 2 1 1 2 2 arctan    + = − − x x x x d 2 2 1 1 1 21 arctan 2 x (x x ) C x = − − arctan + 21 arctan 22 ( ) . 21 1 arctan 21 2 = x + x − x + C 例 6 解