note12 - Math021, week 12 Partial Fractions Example 12.1...

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Math021, week 12 Partial Fractions Example 12.1 Evaluate Z dx x 2 - 1 . solution: Since 1 x 2 - 1 = 1 ( x - 1)( x +1) , we will try finding two numbers A and B such that 1 ( x - 1)( x + 1) = A x - 1 + B x + 1 if possible. If such numbers A and B are there, they satisfy 1 ( x - 1)( x + 1) = ( A + B ) x + ( A - B ) ( x - 1)( x + 1) for all x . That is, we need A + B = 0 and A - B = 1. The choice A = 1 / 2 and B = - 1 / 2 does the job. Thus, 1 ( x - 1)( x + 1) = 1 2 ( 1 x - 1 - 1 x + 1 ) . Finally, R dx x 2 - 1 = 1 2 R ( 1 x - 1 - 1 x +1 ) dx = 1 2 (ln( x - 1) - ln( x + 1)) + C = 1 2 ln x - 1 x +1 + C. Remark 12.2 In the previous example, we split a fraction into a sum of several simpler fractions. Such technique is called the technique of partial fractions. Example 12.3 Evaluate Z sec xdx. 1
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Let u = sin x so that du = cos xdx . Then, R sec xdx = R cos xdx 1 - sin 2 x = R du 1 - u 2 = 1 2 R ( 1 1 - u + 1 1+ u ) du = 1 2 ( - ln(1 - u ) + ln(1 + u )) + C = 1 2 ln 1+sin x 1 - sin x + C Example 12.4 Evaluate Z x 3 dx x 2 - 1 solution: Note that x 3 = x ( x 2 - 1) + x, we have x 3 x 2 - 1 = x ( x 2 - 1) x 2 - 1 + x x 2 - 1 = x + 1 2 x - 1+ x +1 ( x - 1)( x +1) = x + 1 2 ( 1 x - 1 + 1 x +1 ) . Hence, R x 3 dx x 2 - 1 = R [ x + 1 2 ( 1 x - 1 + 1 x +1 )] dx = 1 2 ( x 2 + ln( x - 1) + ln( x + 1)) + C = 1 2 ( x 2 + ln( x 2 - 1)) + C. Example 12.5 Evaluate Z x + 1 x ( x - 1) 3 dx solution: We will try finding three numbers A and B such that x + 1 x ( x - 1) 3 = A x + B ( x - 1) 3 if possible. If such numbers A and B are there, they satisfy x + 1 x ( x - 1) 3 = A ( x - 1) 3 + Bx x ( x - 1) 3 for all x . That is, we need A ( x - 1) 3 + Bx = x + 1 for all x . Putting x = 1 yields B = 2 and putting x = 0 yields A = - 1. But it is obvious that - ( x - 1) 3 + x 6 = x + 1!! We conclude that such numbers
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This note was uploaded on 02/13/2012 for the course MATH 021 taught by Professor Luxnu during the Fall '10 term at HKUST.

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note12 - Math021, week 12 Partial Fractions Example 12.1...

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