Chọn D
Ta có: x 3 - 3 x + 2 = ( x - 1 ) 2 ( x + 2 )
2 x + 3 = a ( x - 1 ) 2 + b ( x + 2 ) ( x - 1 ) + c ( x + 2 ) ⇔ 2 x + 3 = ( a + b ) x 2 + ( c - 2 a + b ) x + a - 2 b + 2 c
⇔ a + b = 0 - 2 a + b + c = 2 a - 2 b + 2 c = 3 ⇔ a = - 1 9 , b = 1 9 , c = 5 3
J = ∫ 2 3 - 1 9 1 x + 2 + 1 9 1 x - 1 + 5 3 1 x - 1 2 d x = 1 9 ln x - 1 x + 2 - 5 3 ( x - 1 ) | 2 3 = 1 9 ln 8 5 + 5 6