∫1/(x^2-2x+3)dx=∫1/[(x-3)(x+1)]dx=∫(1/4){[1/(x-3)]-1/[(x+1)]}dx=(1/4)∫[1/(x-3)]d(x-3)-(1/4)∫1/[(x+1)]}d(x+1)=(1/4)ln|x-3|-(1/4)ln|x+1|+c=(1/4)ln|(x-3)/(x+1)|+c=ln{[(x-3)/(x+1)]^(1/4)}+c
