prob4sol

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

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Math 151 Spring 208 Geveci Problem Set # 4 Integrals of Rational Functions Solutions 1. The expression 4 x 2 +8 x +5 is irreducible since 8 2 4(5)(4)= 16 < 0 . We complete the square: 4 x 2 x +5=4 ¡ x 2 +2 x ¢ +5=4( x +1) 2 4+5=4( x 2 +1 . Therefore, Z 1 4 x 2 x dx = Z 1 4( x 2 dx = Z 1 (2 ( x +1)) 2 dx. We set u =2( x so that du =2 dx .Thu s , Z 1 (2 ( x 2 dx = 1 2 Z 1 u 2 du = 1 2 arctan ( u )= 1 2 arctan (2 x +2) . 2. The expression x 2 x is irreducible since 2 2 4(5)= 16 < 0 .Wecomp letethe square: x 2 x +5=( x 2 +4 . Thus, Z x x 2 x dx = Z x ( x 2 dx = Z x ( x 2 dx Z 1 ( x 2 dx = 1 2 ln ¡ x 2 x ¢ 1 2 arctan μ x 2 3. a) 3 x +16 x 2 + x 12 = 3 x ( x +4)( x 3) = A x + B x 3 3 x +16= A ( x 3) + B ( x +4) . Set x = 4: 28 = 7 A A = 4 . Set x =3: 7=7 B B =1 . Thus, 3 x x 2 + x 12 = 4 x + 1 x 3 1

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b) Z 3 x +16 x 2 + x 12 dx = 4 Z 1 x +4 dx + Z 1 x 3 dx = 4ln( | x | )+ln( | x 3 | ) . 4. a) x +7 ( x 2) ( x 5) = A x 2 + B x 5 x +7= A ( x 5) + B ( x 2) Set x =2: 9= 3 A A = 3 . Set x =5: 12 = 3 B B =4 . Thus, x ( x 2) ( x 5) = 3 x 2 + 4 x 5 b) Z x ( x 2) ( x 5) dx = 3 Z 1 x 2 dx Z 1 x 5 dx = 3ln( | x 2 | )+4ln( | x 5 | ) .
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## This homework help was uploaded on 04/21/2008 for the course MATH 151 taught by Professor Geveci during the Winter '08 term at San Diego State.

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

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