⇒9x+15=A(x2+3x+3)+(Bx+c)(x−3) ⇒9x+15=x2(A+B)+x(3A−3B+C)+(3A−3C) On equating coefficients of x2,x and constant term, we get A+B=0,3A−3B+C=9 and 3A−3C=15 On solving above three equations We get, A=2,B=−2 and C=−3 ∴∫
9x+15
(x−3)(x2+3x+3)
dx =∫
2
x−3
dx+∫
−2x−3
x2+3x+3
dx =∫
2
x−3
dx−∫
2x+3
x2+3x+3
dx =2log|x−3|−log|x2+3x+3|+C ∴A=2,B=−1,g(x)=x−3 and f(x)=x2+3x+3 ∴