Math 151 Spring 208 GeveciProblem Set # 4Integrals of Rational FunctionsSolutions1.The expression4x2+ 8x+ 5is irreducible since82−4 (5) (4) =−16<0.We completethe square:4x2+ 8x+ 5 = 4¡x2+ 2x¢+ 5 = 4 (x+ 1)2−4 + 5 = 4 (x+ 1)2+ 1.Therefore,Z14x2+ 8x+ 5dx=Z14 (x+ 1)2+ 1dx=Z1(2 (x+ 1))2+ 1dx.We setu= 2 (x+ 1)so thatdu= 2dx. Thus,Z1(2 (x+ 1))2+ 1dx=12Z1u2+ 1du=12arctan (u) =12arctan (2x+ 2).2.The expressionx2+ 2x+ 5is irreducible since22−4 (5) =−16<0. We complete thesquare:x2+ 2x+ 5 = (x+ 1)2+ 4.Thus,Zxx2+ 2x+ 5dx=Zx(x+ 1)2+ 4dx=Zx+ 1(x+ 1)2+ 4dx−Z1(x+ 1)2+ 4dx=12ln¡x2+ 2x+ 5¢−12arctanμx+ 12¶3.a)−3x+ 16x2+x−12=−3x+ 16(x+ 4) (x−3)=Ax+ 4+Bx−3⇔−3x+ 16 =A(x−3) +B(x+ 4).Setx=−4 :28 =−7A⇒A=−4.Setx= 3 :7 = 7B⇒B= 1.Thus,−3x+ 16x2+x−12=−4x+ 4+1x−31