例2求 3+2x 解 (3+2x), 3+2x23+2x 3+2x2J3+2x (3+2x)dx d(3+2x) 23+2x du==lnu+C==In 3+2x+C 2 般地f(ax+b)d=Uf(a)dl lu=ax+b例2 求 . 3 2 1 dx x  + 解 (3 2 ) , 3 2 1 2 1 3 2 1  +  + =  + x x x dx x  3 + 2 1 x dx x (3 2 ) 3 2 1 2 1  +  + =  du 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 (3 2 ) 3 2 1 2 1 d x x + + = 