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Unformatted text preview: 8.4 Partial Fractions The Empire Builder, 1957 Greg Kelly, Hanford High School, Richland, Washington 1 5x  3 x 2  2 x  3 dx This would be a lot easier if we could rewrite it as two separate terms. Multiply by the common denominator. 5x  3 A B = + ( x  3) ( x + 1) x  3 x + 1 5 x  3 = A ( x + 1) + B ( x  3)
5 x  3 = Ax + A + Bx  B 3 5x = Ax + Bx 5 = A+ B 3 = A  B 3 3 = A  3B Set liketerms equal to each other. Solve two equations with two unknowns. 1 5x  3 x 2  2 x  3 dx 5 = A+ B 3 =  A + 3B 8 = 4B 3 = A  3B 5x  3 A B = + ( x  3) ( x + 1) x  3 x + 1 5 x  3 = A ( x + 1) + B ( x  3)
5 x  3 = Ax + A + Bx  B 3 5x = Ax + Bx 5 = A+ B 3 = A  B 3 3 = A  3B 2=B 5 = A+ 2 3= A 3 2 x  3 + x + 1 dx
3ln x  3 + 2 ln x + 1 + C
This technique is called Solve two equations with twoPartial Fractions unknowns. Good News!
The AP Exam only requires nonrepeating linear factors! The more complicated methods of partial fractions are good to know, and you might see them in college, but they will not be on the AP exam or on my exam. 2 6x + 7 ( x + 2) 2 A B = + x + 2 ( x + 2) 2 Repeated roots: we must use two terms for partial fractions. 6x + 7 = A ( x + 2) + B 6 x + 7 = Ax + 2 A + B 6x = Ax 6= A 7 = 2A + B 7 = 26 + B 7 = 12 + B 5 = B 6 5  x + 2 ( x + 2) 2 4 2 x3  4 x 2  x  3 x2  2x  3 If the degree of the numerator is higher than the degree of the denominator, use long division first. 2x x 2 2 x 3 2 x3 4 x 2 x 3 2 x3  4 x 2  6 x 5x  3
5x  3 2x + 2 x  2x  3
(from example one) 5x  3 3 2 2x + = 2x + + ( x  3) ( x + 1) ( x  3) ( x + 1) A challenging example: first degree numerator (x 2 x + 4
2 + 1) ( x  1) 2 Ax + B C D = 2 + + x + 1 x  1 ( x  1) 2 irreducible quadratic factor repeated root 2 x + 4 = ( Ax + B ) ( x  1) + C x 2 + 1 ( x  1) + D x 2 + 1
2 ( ) ( ) 2 x + 4 = ( Ax + B ) x 2  2 x + 1 + C x 3  x 2 + x  1 + Dx 2 + D
2 x + 4 = Ax 3  2 Ax 2 + Ax + Bx 2  2 Bx + B + Cx 3  Cx 2 + Cx  C + Dx 2 + D ( ) ( ) 2 x + 4 = Ax 3  2 Ax 2 + Ax + Bx 2  2 Bx + B + Cx 3  Cx 2 + Cx  C + Dx 2 + D 0 = A + C 0 = 2A + B  C + D 2 = A  2 B + C
1 2 1 0 1 0 0 0 0 1 1 1 0 1 0 1 0 1 0 1 0 0 +2 r 3 2  r 1 4 0 4 2 ( 2 ) 4 1 0 0 0 1 0 0 0 0 1 1 0 4 = B C + D
0 0 1 1 0 0 1 1 0 1 4 +3 r 2 4 r2 0 1 1 3 +r3 2 1 1 1 0 1 3 1 1 1 0 1 0 0 1 0 1 1 3 1 2 0 1 1 1 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 0 1 1 0 0 1 1 0 0 1 2 0 1 4 +3 r 2 4 r2 0 1 1 3 +r3 0 1 1 2 2 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 1 0 1 0 1 0 1 0 0 0 1 0 0 0 1 1 0 0 0 1 0 0 0 1 0 1 1  r 4 1 0 r3 1 2 1 2 1 2 1 3 1 1 1 0 1 0 0 0 1 0 0 1 0 1 1 1 0 1 0 (x 2 x + 4
2 + 1) ( x  1) 2 Ax + B C D = 2 + + x + 1 x  1 ( x  1) 2 2x +1 2 1 = 2  + 2 x + 1 x  1 ( x  1) We can do this problem on the TI89: 2 x + 4 expand 2 2 ( x + 1) ( x  1) expand ((2x+4)/((x^2+1)*(x1)^2)) 1 0 0 0 2 0 1 0 0 1 F2 3 0 0 1 0 2 2 x 1 2 1 Of course with the TI89, we could + 2  + 0 0 1 1 just integrate 0 and wouldn't need x 2 + 1 x + 1 x  1 ( x  1) 2
partial fractions! ...
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 Fall '05
 Riggs
 Fractions, Differential Calculus

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