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5. } x 2 2 x 2 5 1 x 8 1 6 } 5 1 1 } x 2 5 2 x 5 1 x 2 1 6 } x 2 2 5 x 1 6 5 ( x 2 3)( x 2 2) } x 2 5 2 x 5 1 x 2 1 6 } 5 } x 2 A 3 } 1 } x 2 B 2 } 5 x 1 2 5 A ( x 2 2) 1 B ( x 2 3) 5 ( A 1 B ) x 1 ( 2 2 A 2 3 B ) Equating coefficients of like terms gives A 1 B 5 5 and 2 2 A 2 3 B 5 2 Solving the system simultaneously yields A 5 17, B 5 2 12. } x 2 2 x 2 5 1 x 8 1 6 } 5 1 1 } x 1 2 7 3 } 2 } x 1 2 2 2 } 6. } y y 3 2 1 1 1 4 } 5 y 1 } 2 y 4 2 y 1 1 4 1 } The rational function cannot be decomposed any further. 7. 1 2 x 2 5 (1 2 x )(1 1 x ) } 1 2 1 x 2 } 5 } 1 2 A x } 1 } 1 1 B x } 1 5 A (1 1 x ) 1 B (1 2 x ) 5 ( A 2 B ) x 1 ( A 1 B ) Equating coefficients of like terms gives A 2 B 5 0 and A 1 B 5 1. Solving the system simultaneously yields A 5 } 1 2 } , B 5 } 1 2 } . E } 1 2 dx x 2 } 5 E } 1 1 2 /2 x } dx 1 E } 1 1 1 /2 x } dx 5 2 } 1 2 } ln ) 1 2 x ) 1 } 1 2 } ln ) 1 1 x ) 1 C 5 } 1 2 } ln ) } 1 1 1 2 x x } ) 1 C 8. x 2 1 2 x 5 x ( x 1 2) } x 2 1 1 2 x } 5 } A x } 1 } x 1 B 2 } 1 5 A ( x 1 2) 1 Bx 5 ( A 1 B ) x 1 2 A Equating coefficients of like terms gives A 1 B 5 0 and 2 A 5 1 Solving the system simultaneously yields A 5 } 1 2 } , B 5 2 } 1 2 } . E } x 2 d 1 x 2 x } 5 E } 1 x /2 } dx 1 E } x 2 1 1/2 2 } dx 5 } 1 2 } ln ) x ) 2 } 1 2 } ln ) x 1 2 ) 1 C 5 } 1 2 } ln ) } x 1 x 2 } ) 1 C 9. y 2 2 2 y 2 3 5 ( y 2 3)( y 1 1) } y 2 2 2 y y 2 3 } 5 } y 2 A 3 } 1 } y 1 B 1 } y 5 A ( y 1 1) 1 B ( y 2 3) 5 ( A 1 B ) y 1 ( A 2 3 B ) Equating coefficients of like terms gives A 1 B 5 1 and A 2 3 B 5 0. Solving the system simultaneously yields A 5 } 3 4 } , B 5 } 1 4 } . E } y 2 2 y 2 d y y 2 3 } 5 E } y 3 2 /4 3 } dy 1 E } y 1 1 /4 1 } dy 5 } 3 4 } ln ) y 2 3 ) 1 } 1 4 } ln ) y 1 1 ) 1 C 10. y 2 1 y 5 y ( y 1 1) } y y 2 1 1 4 y } 5 } A y } 1 } y 1 B 1 } y 1 4 5 A ( y 1 1) 1 By 5 ( A 1 B ) y 1 A Equating coefficients of like terms gives A 1 B 5 1 and A 5 4. Solving the system simultaneously yields A 5 4, B 5 2 3. E } y y 2 1 1 4 y } dy 5 E } 4 y } dy 1 E } y 2 1 3 1 } dy 5 4 ln ) y ) 2 3 ln ) y 1 1 ) 1 C y y 2 1 4 q y 3 w 1 w 0 w y 2 w 1 w 0 w y w 1 w 1 w y 3 1 0 y 2 1 4 y 1 0 }}}}}}} 2 4 y 1 1 1 x 2 2 5 x 1 6 q x 2 w 1 w 0 w x w 1 w 8 w x 2 2 5 x 1 6 }}}} 5 x 1 2 346 Section 8.4

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11. t 3 1 t 2 2 2 t 5 t ( t 2 1 t 2 2) 5 t ( t 1 2)( t 2 1) } t 3 1 t 1 2 2 2 t } 5 } A t } 1 } t 1 B 2 } 1 } t 2 C 1 } 1 5 A ( t 1 2)( t 2 1) 1 B ( t )( t 2 1) 1 C ( t )( t 1 2) 5 A ( t 2 1 t 2 2) 1 B ( t 2 2 t ) 1 C ( t 2 1 2 t ) 5 ( A 1 B 1 C ) t 2 1 ( A 2 B 1 2 C ) t 2 2 A Equating coefficients of like terms gives A 1 B 1 C 5 0, A 2 B 1 2 C 5 0, and 2 2 A 5 1. Solving the system simultaneously yields A 5 2 } 1 2 } , B 5 } 1 6 } , C 5 } 1 3 } . E } t 3 1 d t 2 t 2 2 t } 5 E } 2 1 t /2 } dt 1 E } t 1 1 /6 2 } dt 1 E } t 1 2 /3 1 } dt 5 2 } 1 2 } ln ) t ) 1 } 1 6 } ln ) t 1 2 ) 1 } 1 3 } ln ) t 2 1 ) 1 C 12. 2 t 3 2 8 t 5 2 t ( t 2 2 4) 5 2 t ( t 2 2)( t 1 2) } 2 t t 3 1 2 3 8 t } 5 } A t } 1 } t 2 B 2 } 1 } t 1 C 2 } t 1 3 5 2 A ( t 2 2)( t 1 2) 1 2 B ( t )( t 1 2) 1 2 C ( t )( t 2 2) 5 2 A ( t 2 2 4) 1 2 B ( t 2 1 2 t ) 1 2 C ( t 2 2 2 t ) 5 (2 A 1 2 B 1 2 C ) t 2 1 (4 B 2 4 C ) t 2 8 A Equating coefficients of like terms gives 2 A 1 2 B 1 2 C 5 0, 4 B 2 4 C 5 1, 2 8 A 5 3 Solving the system simultaneously yields A 5 2 } 3 8 } , B 5 } 1 5 6 } , C 5 } 1 1 6 } .
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