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Example 3 Evaluate

Example 3 Evaluate

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Example 3 Evaluate

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  1. Example 3 Evaluate Solution Since the degree 5 of the numerator is less than the degree 6 of the denominator we do not need to do long division. The denominator is given in factored form and neither x2+4x+29 nor x2+2x+10 has any real roots. We will use the method of partial fractions to write Begin by adding the fractions on the right and equating the numerators: We use the first variation of the method of partial fractions. We evaluate the products on the right and collect the coefficients of each power of x.

  2. Equate the coefficients of each power of x: x5: 2 = A + E x4: 12 = 6A + B + 8E + F x3: 138 = 47A + 6B + C +74E + 8F x2: 210= 98A + 47B + 2C + D + 232E + 74F x1: 1272 = 290A + 98B + 10C + 2D + 841E + 232F x0: -5065 = 290B + 10D + 841F

  3. We will show in an appendix to this example that the unique solution to the preceding linear system of 6 equations in 6 unknowns is: A=-2, B=-3, C=-6, D=1, E=4, F=-5. In the first two integrals substitute x+2 = 5tan  with dx = 5sec2  d while in the third integral substitute x+1 = 3tan  with dx = 3sec2 d: Then

  4. Since tan  = (x+2)/5, we have  = arctan (x+2)/5. Since tan  = (x+1)/3, we have  = arctan (x+1)/3. From the right triangles below: Then  

  5. Appendix Solve the following linear system: From equations (1) : E = 2 - A. By equation (2): 12 = 6A + B + 8(2 - A) + F = 16 - 2A + B + F and F = 2A – B - 4. Substitute these values of E and F into the other equations: 138 = 47A + 6B + C +74(2-A) + 8(2A-B-4) (3a) 210= 98A + 47B + 2C + D + 232(2-A) + 74(2A-B-4)(4a) 1272 = 290A + 98B + 10C + 2D + 841(2-A) + 232(2A-B-4)(5a) -5065 = 290B + 10D + 841(2A-B-4)(6a) 2 = A + E (1) 12 = 6A + B + 8E + F (2) 138 = 47A + 6B + C +74E + 8F (3) 210= 98A + 47B + 2C + D + 232E + 74F (4) 1272 = 290A + 98B + 10C + 2D + 841E + 232F (5) -5065 = 290B + 10D + 841F (6)

  6. Hence 22 = -11A - 2B + C (3b) 42 = 14A - 27B + 2C + D (4b) 518 = -87A - 134B + 10C + 2D (5a) -1701 = 1682A - 551B + 10D (6a) From equation (3b): C = 22 + 11A + 2B. Substitute this value of C into the equations (4b) and (5a): 42 = 14A - 27B + 2(22 + 11A + 2B) + D (4c) 518 = -87A - 134B + 10 (22 + 11A + 2B) + 2D (5b) -1701 = 1682A - 551B + 10D (6a) Then -2 = 36A - 23B + D (4d) 298 = 23A - 114B + 2D (5c) -1701 = 1682A - 551B + 10D (6a) From equation (4d): D = -36A + 23B – 2.

  7. D = -36A+23B-2 Substitute this value of D into equations (5c) and (6a): 298 = 23A - 114B + 2(-36A + 23B - 2) (5d) -1701 = 1682A - 551B + 10(-36A + 23B - 2)(6b) Then 302 = -49A - 68B(5e) 1681 = -1322A + 321B (6c) Add 321 times equation (5e) with 68 times equation (6c): 321(302) + 68(1681) = 321(-49)A + 68(-1322A) 211,250 = -105,625A and A=-2. By equation (5e): 302 = -49(-2) - 68B = 98 – 68B, so 68B = -204 and B=-3. From the boxed equation above: D = -36A + 23B - 2 = -36(-2) + 23(-3) - 2 = 72 – 69 - 2 = 1. By equation (3b): C = 22 + 11A +2B = 22 + 11(-2) + 2(-3) = -6. From page 5: E = 2 - A = 2 – (-2) = 4 while F = 2A – B - 4 = 2(-2) – (-3) – 4 = -5. Thus A=-2, B=-3, C=-6, D=1, E=4, F=-5.

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