(6)∫f((tanx)scc2xdr=∫f(tanx)dtanx (7)∫fe)e'dr=∫fe)de (图)J/Qnxax=-jrh)dnx s求厂u2n时 dx 解：限式-inx02四 1+2x+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8 求  1+ 2ln x dln x 解: 原式 =  + = 2 1 2ln x 1 d(1+ 2ln x)