∫[1/(x^2-x-6)] dx=∫[1/(x-3)(x+2)] dx=∫(1/5)[1/(x-3) -1/(x+2)] dx=(1/5)∫[1/(x-3) -1/(x+2)]dx=(1/5)[∫1/(x-3) d(x-3) -∫1/(x+2) d(x+2)]=(1/5)(ln|x-3|-ln|x+2|) +C