prob4sol - Math 151 Spring 208 Geveci Problem Set # 4...

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Math 151 Spring 208 GeveciProblem Set # 4Integrals of Rational FunctionsSolutions1.The expression4x2+ 8x+ 5is irreducible since824 (5) (4) =16<0.We completethe square:4x2+ 8x+ 5 = 4¡x2+ 2x¢+ 5 = 4 (x+ 1)24 + 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 since224 (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+ 4dxZ1(x+ 1)2+ 4dx=12ln¡x2+ 2x+ 5¢12arctanμx+ 123.a)3x+ 16x2+x12=3x+ 16(x+ 4) (x3)=Ax+ 4+Bx33x+ 16 =A(x3) +B(x+ 4).Setx=4 :28 =7AA=4.Setx= 3 :7 = 7BB= 1.Thus,3x+ 16x2+x12=4x+ 4+1x31
b)Z3x+ 16x2+x12dx=4Z1x+ 4dx+Z1x3dx=4 ln (|x+ 4|) + ln (|x3|).4.a)x+ 7(x2) (x5)=Ax2+Bx5x+ 7 =A(x5) +B(x2)Setx= 2 :9 =3AA=3.Setx= 5 :12 = 3BB= 4.Thus,x+ 7(x2) (x5)=3x2+4x5b)Zx+ 7(x2) (x5)dx=3Z1x2dx+ 4Z1x5dx=3 ln (|x2|) + 4 ln (|x5|).

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Term
Winter
Professor
Geveci
Tags
Integrals, Rational Functions, dx, Halle Berry

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