例3求积分 arctan xd 2 解令u= arctan,xdhc=dx=bh 2 ∫ x arctan xdx=; arctan-「,a d(arctan) 2 2 arctan 2 21+x 2 =-arctanx Ddx 2 2 2arctanx-2x-arctanx)+c例3 求积分 arctan .  x xdx 解 令 u = arctan x , dv x xdx = d = 2 2  xarctan xdx (arctan ) 2 arctan 2 2 2 d x x x x  = − dx x x x x 2 2 2 1 1 2 arctan 2 + = −   dx x x x ) 1 1 (1 2 1 arctan 2 2 2 + = −  −  ( arctan ) . 2 1 arctan 2 2 x x x C x = − − +