Find the integral of ((x^2)/(x^2+7x+12)) with respect to x

x^2/(x+3)(x+4)=1+A/(x+3)+B/(x+4) and then L.c.m, and cancel LCM on both sides the remaining is. x^2=1+A(x+4)+B(x+3)....…...(1)put x=-3 on both sides we getA=9,put x=-4 getB=-16Therefore integral ofx^2/(x+3)(x+4)=x+9 log(x+3)-16log(x+4)