例3求 x(1+2In x) 解:原式 -d(n x) 1+2Inx 2J1+2Inx -d(+2In x 1(令v=1+lnx) 21212 =In u+c =1l|1+2mx|+C例3 求 dx x x  (1+ 2ln ) 1 解: 原式= (ln ) 1 2ln 1 d x x  + (1 2ln ) 1 2ln 1 2 1 d x x + + =  du (令u=1+2lnx) u  = 1 2 1 = ln | u | +C 2 1 = ln |1+ 2ln x | +C 2 1