121616949-math.187 - that the numerator has degree less...

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8.5 Rational Functions 173 If we go ahead and add the fractions on the right hand side we get 7 x - 6 ( x - 2)( x + 3) = ( A + B ) x + 3 A - 2 B ( x - 2)( x + 3) . So all we need to do is find A and B so that 7 x - 6 = ( A + B ) x + 3 A - 2 B , which is to say, we need 7 = A + B and - 6 = 3 A - 2 B . This is a problem you’ve seen before: solve a system of two equations in two unknowns. There are many ways to proceed; here’s one: If 7 = A + B then B = 7 - A and so - 6 = 3 A - 2 B = 3 A - 2(7 - A ) = 3 A - 14+2 A = 5 A - 14. This is easy to solve for A : A = 8 / 5, and then B = 7 - A = 7 - 8 / 5 = 27 / 5. Thus Z 7 x - 6 ( x - 2)( x + 3) dx = Z 8 5 1 x - 2 + 27 5 1 x + 3 dx = 8 5 ln | x - 2 | + 27 5 ln | x + 3 | + C. The answer to the original problem is now Z x 3 ( x - 2)( x + 3) dx = Z x - 1 dx + Z 7 x - 6 ( x - 2)( x + 3) dx = x 2 2 - x + 8 5 ln | x - 2 | + 27 5 ln | x + 3 | + C. Now suppose that x 2 + bx + c doesn’t factor. Again we can use long division to ensure
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Unformatted text preview: that the numerator has degree less than 2. Now we can complete the square to turn the integral into a trigonometric substitution problem. EXAMPLE 8.20 Evaluate Z x x 2 + x + 1 dx . We have seen that this quadratic does not factor. We complete the square: x 2 + x + 1 = x 2 + x + 1 4 + 1-1 4 = x 2 + x + 1 4 + 3 4 = ± x + 1 2 ¶ 2 + 3 4 . Now factor out 3 / 4: 3 4 ˆ 4 3 ± x + 1 2 ¶ 2 + 1 ! = 3 4 ˆ ± 2 √ 3 x + 1 √ 3 ¶ 2 + 1 ! . Now let tan u = 2 √ 3 x + 1 √ 3 sec 2 u du = 2 √ 3 dx dx = √ 3 2 sec 2 u du....
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