d=lnu+C、 ln(3+2x)+C 2 般地∫f(ax+b)tx=(a)hl u=ax+b 例3求 x(1+2Inx) 解 dh x(1+2nx) d(nx) 1+2Inx d(1+2Inx 2J1+2Inx u=1+2lnx 2 -du =Inu+C=In(1+2In x)+C. 2du u  = 1 2 1 = lnu + C 2 1 ln(3 2 ) . 2 1 = + x + C  f (ax + b)dx =  u du u=ax+b f a [ ( ) ] 1 一般地 例3 求 . (1 2ln ) 1 dx x x  + 解 dx x x  (1+ 2ln ) 1 (ln ) 1 2ln 1 d x x  + = (1 2ln ) 1 2ln 1 2 1 d x x + + =  u = 1+ 2ln x  = du u 1 2 1 = lnu + C 2 1 ln(1 2ln ) . 2 1 = + x + C