∫1/(2-3x)dx=(-1/3)∫1/(2-3x)d(2-3x).因为d(2-3x)/dx=-3=-1/3ln|2-3x|+C;∫1/(X^2-x-12)dx=∫1/(x-4)(x+3)dx=(1/7)∫[x+3-(x-4)]/(x-4)(x+3)dx=(1/7)[∫1/(x-4)dx-∫1/(x+3)dx]=(1/7)(ln|x-4|-ln|x+3|)+C=(1/7)ln|...