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