答：1/[(1-x)x^2)]=a/(1-x)+(bx+c)/x^2=(ax^2+bx+c-bx^2-cx)/[(1-x)x^2]a-b=0b-c=0c=1解得：a=b=c=1∫{1/[(1-x)x^2]}dx=∫[1/(1-x)]dx+∫[(x+1)/x^2]dx=-ln|1-x|+ln|x|-1/x+C=ln|x/(1-x)|-1/x+C