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1 Example 4 Evaluate Solution Since the degree 7 of the numerator is less than the degree 8 of the denominator we do not need to do long division. The.

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Presentation on theme: "1 Example 4 Evaluate Solution Since the degree 7 of the numerator is less than the degree 8 of the denominator we do not need to do long division. The."— Presentation transcript:

1 1 Example 4 Evaluate Solution Since the degree 7 of the numerator is less than the degree 8 of the denominator we do not need to do long division. The denominator factors as: x 8 +3x 6 +3x 4 +x 2 = x 2 (x 6 +3x 4 +3x 2 +1) = x 2 (x 2 +1) 3 and x 2 +1 has no 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 7 : 9 = A + C (1) x 6 : -10 = B + D (2) x 5 : 17 = 3A + 2C + E (3) x 4 : -26 = 3B + 2D + F (4) x 3 : 9 = 3A + C + E + G (5) x 2 : -28 = 3B + D + F + H (6) x 1 : 3 = A (7) x 0 : -7 = B (8) From equation 1: 9 = 3 + C and C = 6. From equation 2: -10 = -7 + D and D = -3. From equation 3: 17 = 9 + 12 + E and E = -4. From equation 4: -26 = -21 - 6 + F and F = 1. From equation 5: 9 = 9 + 6 – 4 + G and G = -2 From equation 6: -28 = -21 – 3 + 1 + H and H = -5.

3 3 A=3, B=-7, C=6, D = -3, E = -4, F=1, G=-2, H = -5. Then Substitute x = tan  with dx = sec 2  d  in all three integrals:

4 4 Since tan  = x, we have  = arctan x. From the right triangle below: 

5 5 Then


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