1. 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 x 2 +4x+29 nor x 2 +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. 2 Equate the coefficients of each power of x: x 5 : 2 = A + E x 4 : 12 = 6A + B + 8E + F x 3 : 138 = 47A + 6B + C +74E + 8F x 2 : 210= 98A + 47B + 2C + D + 232E + 74F x 1 : 1272 = 290A + 98B + 10C + 2D + 841E + 232F x 0 : -5065 = 290B + 10D + 841F

3. 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 = 5sec 2  d  while in the third integral substitute x+1 = 3tan  with dx = 3sec 2  d  : Then

4. 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. 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. 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. 7 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. D = -36A+23B-2