(6)f(tanx)sec xdx=I f(tan x)dtanx (7)If(e )e dx= f(e")de )∫mx)dx=J/(mlnx dx 例3求」 r(1+2Inx) 解:原式= dInx 1 rd(1+2Inx' 1+2Inx 2J1+2Inx In 1+2Inx+C=  (6) f (tan x)sec xdx 2 dtan x =  f e e x x x (7) ( ) d x de =  x x f x d 1 (8) (ln ) dln x 例3. 求  1+ 2ln x dln x 解: 原式 =  + = 2 1 2ln x 1 d(1+ 2ln x)