∫（x^2-3x+2）^3（2x-3）dx =∫（x^2-3x+2）^3d（x^2-3x+2）=[（x^2-3x+2）^4]/4 +C
解:∫(x²-3x+2)³(2x-3) dx =∫(x-1)³(x-2)³(2x-3) dx
