原式=∫[1/(x-1)-1/x]dx
=ln|x-1|-ln|x|+C
=ln|(x-1)/x|+C原式是平方分之一a/x+b/(x-1)+c/(x-1)²=1/x(x-1)²a=1,b=-1,c=-3所以原式=ln|x|-ln|x-1|+3/(x+1)+C